Japanese Temple Problem From 1844
$begingroup$
I recently learnt a Japanese geometry temple problem.
The problem is the following:
Five squares are arranged as the image shows. Prove that the area of triangle T and the area of square S are equal.
This is problem 6 in this article.
I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated.
geometry sangaku
edited Nov 22 '18 at 20:49
Jean-Claude Arbaut
14.8k63464
asked Nov 22 '18 at 20:40
LarryLarry
2,39131129
$endgroup$
add a comment |
$begingroup$
I recently learnt a Japanese geometry temple problem.
The problem is the following:
Five squares are arranged as the image shows. Prove that the area of triangle T and the area of square S are equal.
This is problem 6 in this article.
I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated.
geometry sangaku
edited Nov 22 '18 at 20:49
Jean-Claude Arbaut
14.8k63464
asked Nov 22 '18 at 20:40
LarryLarry
2,39131129
$endgroup$
3
$begingroup$
If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
$endgroup$
– mickep
Nov 24 '18 at 14:03
$begingroup$
I checked it out. A great book indeed, thank you.
$endgroup$
– Larry
Nov 24 '18 at 14:22
$begingroup$
The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:24
1
$begingroup$
^ From J.G.'s answer.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:26
add a comment |
$begingroup$
I recently learnt a Japanese geometry temple problem.
The problem is the following:
Five squares are arranged as the image shows. Prove that the area of triangle T and the area of square S are equal.
This is problem 6 in this article.
I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated.
geometry sangaku
edited Nov 22 '18 at 20:49
Jean-Claude Arbaut
14.8k63464
asked Nov 22 '18 at 20:40
LarryLarry
2,39131129
$endgroup$
I recently learnt a Japanese geometry temple problem.
The problem is the following:
Five squares are arranged as the image shows. Prove that the area of triangle T and the area of square S are equal.
This is problem 6 in this article.
I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated.
geometry sangaku
geometry sangaku
edited Nov 22 '18 at 20:49
Jean-Claude Arbaut
14.8k63464
asked Nov 22 '18 at 20:40
LarryLarry
2,39131129
edited Nov 22 '18 at 20:49
Jean-Claude Arbaut
14.8k63464
asked Nov 22 '18 at 20:40
LarryLarry
2,39131129
edited Nov 22 '18 at 20:49
Jean-Claude Arbaut
14.8k63464
edited Nov 22 '18 at 20:49
Jean-Claude Arbaut
14.8k63464
edited Nov 22 '18 at 20:49
Jean-Claude Arbaut
14.8k63464
14.8k63464
asked Nov 22 '18 at 20:40
LarryLarry
2,39131129
asked Nov 22 '18 at 20:40
LarryLarry
2,39131129
asked Nov 22 '18 at 20:40
LarryLarry
2,39131129
2,39131129
3
$begingroup$
If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
$endgroup$
– mickep
Nov 24 '18 at 14:03
$begingroup$
I checked it out. A great book indeed, thank you.
$endgroup$
– Larry
Nov 24 '18 at 14:22
$begingroup$
The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:24
1
$begingroup$
^ From J.G.'s answer.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:26
add a comment |
3
$begingroup$
If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
$endgroup$
– mickep
Nov 24 '18 at 14:03
$begingroup$
I checked it out. A great book indeed, thank you.
$endgroup$
– Larry
Nov 24 '18 at 14:22
$begingroup$
The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:24
1
$begingroup$
^ From J.G.'s answer.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:26
3
3
$begingroup$
If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
$endgroup$
– mickep
Nov 24 '18 at 14:03
$begingroup$
If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
$endgroup$
– mickep
Nov 24 '18 at 14:03
$begingroup$
I checked it out. A great book indeed, thank you.
$endgroup$
– Larry
Nov 24 '18 at 14:22
$begingroup$
I checked it out. A great book indeed, thank you.
$endgroup$
– Larry
Nov 24 '18 at 14:22
$begingroup$
The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:24
$begingroup$
The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:24
1
1
$begingroup$
^ From J.G.'s answer.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:26
$begingroup$
^ From J.G.'s answer.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:26
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
We will, first of all, prove a very interesting property
$mathbf{Lemma;1}$
Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.
$mathbf {Proof}$
Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
$$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$
Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$
Now, back to the problem
Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
See why? $mathbf {Hint:}$
It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.
Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$
Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
By Lemma 1:
$$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
The area of the polygon AJKGD is thus
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$
The area of the trapezoid AJKD is moreover
$$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$
Finally
$$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$
edited Nov 23 '18 at 7:43
Mutantoe
610513
answered Nov 22 '18 at 21:39
Dr. MathvaDr. Mathva
1,098317
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Nov 24 '18 at 13:16
add a comment |
$begingroup$
$$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$
$$S ;=; R - 4cdotfrac12ab ;=; T$$
(This space intentionally left blank.)
answered Nov 23 '18 at 3:39
BlueBlue
48.2k870153
$endgroup$
3
$begingroup$
I wonder what tools you use to create awesome graphs like this
$endgroup$
– Larry
Nov 23 '18 at 14:24
5
$begingroup$
@Larry: I use GeoGebra.
$endgroup$
– Blue
Nov 23 '18 at 19:47
1
$begingroup$
I see, thank you.
$endgroup$
– Larry
Nov 23 '18 at 19:49
1
$begingroup$
@crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
$endgroup$
– Blue
Nov 25 '18 at 11:40
1
$begingroup$
@Blue Thanks a lot for clarifying :)
$endgroup$
– crskhr
Nov 25 '18 at 14:34
|
show 1 more comment
$begingroup$
Because there are so many squares, coordinates are easy to compute.
The area of the shaded square is clearly $u^2+v^2$.
The area of the shaded triangle is one-half of the absolute value of the determinant of the array
$$left[ begin{array}{c}
1 & 1 & 1 \
2u-v & 3u & 2u \
3u+v & u+3v & u+v
end{array} right]$$
which is also $u^2+v^2$.
I have a second solution.
$triangle GPN$ is obtained by rotating $triangle GSD 90^circ$ clockwise.
$triangle GQM$ is obtained by rotating $triangle GRK 90^circ$ counterclockwise.
answered Nov 23 '18 at 14:50
steven gregorysteven gregory
18k32258
$endgroup$
add a comment |
$begingroup$
The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.
What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.
answered Nov 24 '18 at 9:25
Will OrrickWill Orrick
13.6k13360
$endgroup$
add a comment |
$begingroup$
While the other solutions are obviously correct, they are also unnecessarily complicated.
Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.
*) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.
While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct
answered Nov 23 '18 at 19:06
DenDenDoDenDenDo
58948
$endgroup$
7
$begingroup$
That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
$endgroup$
– Owen
Nov 23 '18 at 23:19
6
$begingroup$
Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
$endgroup$
– IanF1
Nov 24 '18 at 8:01
1
$begingroup$
I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
$endgroup$
– user3445853
Nov 24 '18 at 20:38
2
$begingroup$
If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
$endgroup$
– user3445853
Nov 24 '18 at 20:40
1
$begingroup$
The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
$endgroup$
– Blue
Dec 16 '18 at 19:53
add a comment |
$begingroup$
This is a long comment.
The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.
The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
answered Nov 22 '18 at 21:11
J.G.J.G.
25.6k22539
$endgroup$
$begingroup$
I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:35
$begingroup$
$BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
$endgroup$
– I like Serena
Nov 22 '18 at 21:41
$begingroup$
$BACD$ is not a rhombus.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:42
2
$begingroup$
@Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
$endgroup$
– J.G.
Nov 23 '18 at 10:22
1
$begingroup$
I have converted your answer into a comment.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:28
|
show 4 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009635%2fjapanese-temple-problem-from-1844%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We will, first of all, prove a very interesting property
$mathbf{Lemma;1}$
Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.
$mathbf {Proof}$
Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
$$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$
Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$
Now, back to the problem
Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
See why? $mathbf {Hint:}$
It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.
Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$
Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
By Lemma 1:
$$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
The area of the polygon AJKGD is thus
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$
The area of the trapezoid AJKD is moreover
$$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$
Finally
$$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$
edited Nov 23 '18 at 7:43
Mutantoe
610513
answered Nov 22 '18 at 21:39
Dr. MathvaDr. Mathva
1,098317
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Nov 24 '18 at 13:16
add a comment |
$begingroup$
We will, first of all, prove a very interesting property
$mathbf{Lemma;1}$
Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.
$mathbf {Proof}$
Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
$$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$
Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$
Now, back to the problem
Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
See why? $mathbf {Hint:}$
It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.
Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$
Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
By Lemma 1:
$$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
The area of the polygon AJKGD is thus
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$
The area of the trapezoid AJKD is moreover
$$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$
Finally
$$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$
edited Nov 23 '18 at 7:43
Mutantoe
610513
answered Nov 22 '18 at 21:39
Dr. MathvaDr. Mathva
1,098317
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Nov 24 '18 at 13:16
add a comment |
$begingroup$
We will, first of all, prove a very interesting property
$mathbf{Lemma;1}$
Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.
$mathbf {Proof}$
Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
$$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$
Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$
Now, back to the problem
Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
See why? $mathbf {Hint:}$
It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.
Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$
Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
By Lemma 1:
$$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
The area of the polygon AJKGD is thus
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$
The area of the trapezoid AJKD is moreover
$$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$
Finally
$$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$
edited Nov 23 '18 at 7:43
Mutantoe
610513
answered Nov 22 '18 at 21:39
Dr. MathvaDr. Mathva
1,098317
$endgroup$
We will, first of all, prove a very interesting property
$mathbf{Lemma;1}$
Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.
$mathbf {Proof}$
Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
$$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$
Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$
Now, back to the problem
Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
See why? $mathbf {Hint:}$
It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.
Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$
Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
By Lemma 1:
$$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
The area of the polygon AJKGD is thus
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$
The area of the trapezoid AJKD is moreover
$$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$
Finally
$$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$
edited Nov 23 '18 at 7:43
Mutantoe
610513
answered Nov 22 '18 at 21:39
Dr. MathvaDr. Mathva
1,098317
edited Nov 23 '18 at 7:43
Mutantoe
610513
edited Nov 23 '18 at 7:43
Mutantoe
610513
edited Nov 23 '18 at 7:43
Mutantoe
610513
610513
answered Nov 22 '18 at 21:39
Dr. MathvaDr. Mathva
1,098317
answered Nov 22 '18 at 21:39
Dr. MathvaDr. Mathva
1,098317
answered Nov 22 '18 at 21:39
Dr. MathvaDr. Mathva
1,098317
1,098317
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Nov 24 '18 at 13:16
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Nov 24 '18 at 13:16
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Nov 24 '18 at 13:16
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Nov 24 '18 at 13:16
add a comment |
$begingroup$
$$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$
$$S ;=; R - 4cdotfrac12ab ;=; T$$
(This space intentionally left blank.)
answered Nov 23 '18 at 3:39
BlueBlue
48.2k870153
$endgroup$
3
$begingroup$
I wonder what tools you use to create awesome graphs like this
$endgroup$
– Larry
Nov 23 '18 at 14:24
5
$begingroup$
@Larry: I use GeoGebra.
$endgroup$
– Blue
Nov 23 '18 at 19:47
1
$begingroup$
I see, thank you.
$endgroup$
– Larry
Nov 23 '18 at 19:49
1
$begingroup$
@crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
$endgroup$
– Blue
Nov 25 '18 at 11:40
1
$begingroup$
@Blue Thanks a lot for clarifying :)
$endgroup$
– crskhr
Nov 25 '18 at 14:34
|
show 1 more comment
$begingroup$
$$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$
$$S ;=; R - 4cdotfrac12ab ;=; T$$
(This space intentionally left blank.)
answered Nov 23 '18 at 3:39
BlueBlue
48.2k870153
$endgroup$
3
$begingroup$
I wonder what tools you use to create awesome graphs like this
$endgroup$
– Larry
Nov 23 '18 at 14:24
5
$begingroup$
@Larry: I use GeoGebra.
$endgroup$
– Blue
Nov 23 '18 at 19:47
1
$begingroup$
I see, thank you.
$endgroup$
– Larry
Nov 23 '18 at 19:49
1
$begingroup$
@crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
$endgroup$
– Blue
Nov 25 '18 at 11:40
1
$begingroup$
@Blue Thanks a lot for clarifying :)
$endgroup$
– crskhr
Nov 25 '18 at 14:34
|
show 1 more comment
$begingroup$
$$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$
$$S ;=; R - 4cdotfrac12ab ;=; T$$
(This space intentionally left blank.)
answered Nov 23 '18 at 3:39
BlueBlue
48.2k870153
$endgroup$
$$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$
$$S ;=; R - 4cdotfrac12ab ;=; T$$
(This space intentionally left blank.)
answered Nov 23 '18 at 3:39
BlueBlue
48.2k870153
edited Nov 23 '18 at 6:02
answered Nov 23 '18 at 3:39
BlueBlue
48.2k870153
answered Nov 23 '18 at 3:39
BlueBlue
48.2k870153
answered Nov 23 '18 at 3:39
BlueBlue
48.2k870153
48.2k870153
3
$begingroup$
I wonder what tools you use to create awesome graphs like this
$endgroup$
– Larry
Nov 23 '18 at 14:24
5
$begingroup$
@Larry: I use GeoGebra.
$endgroup$
– Blue
Nov 23 '18 at 19:47
1
$begingroup$
I see, thank you.
$endgroup$
– Larry
Nov 23 '18 at 19:49
1
$begingroup$
@crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
$endgroup$
– Blue
Nov 25 '18 at 11:40
1
$begingroup$
@Blue Thanks a lot for clarifying :)
$endgroup$
– crskhr
Nov 25 '18 at 14:34
|
show 1 more comment
3
$begingroup$
I wonder what tools you use to create awesome graphs like this
$endgroup$
– Larry
Nov 23 '18 at 14:24
5
$begingroup$
@Larry: I use GeoGebra.
$endgroup$
– Blue
Nov 23 '18 at 19:47
1
$begingroup$
I see, thank you.
$endgroup$
– Larry
Nov 23 '18 at 19:49
1
$begingroup$
@crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
$endgroup$
– Blue
Nov 25 '18 at 11:40
1
$begingroup$
@Blue Thanks a lot for clarifying :)
$endgroup$
– crskhr
Nov 25 '18 at 14:34
3
3
$begingroup$
I wonder what tools you use to create awesome graphs like this
$endgroup$
– Larry
Nov 23 '18 at 14:24
$begingroup$
I wonder what tools you use to create awesome graphs like this
$endgroup$
– Larry
Nov 23 '18 at 14:24
5
5
$begingroup$
@Larry: I use GeoGebra.
$endgroup$
– Blue
Nov 23 '18 at 19:47
$begingroup$
@Larry: I use GeoGebra.
$endgroup$
– Blue
Nov 23 '18 at 19:47
1
1
$begingroup$
I see, thank you.
$endgroup$
– Larry
Nov 23 '18 at 19:49
$begingroup$
I see, thank you.
$endgroup$
– Larry
Nov 23 '18 at 19:49
1
1
$begingroup$
@crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
$endgroup$
– Blue
Nov 25 '18 at 11:40
$begingroup$
@crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
$endgroup$
– Blue
Nov 25 '18 at 11:40
1
1
$begingroup$
@Blue Thanks a lot for clarifying :)
$endgroup$
– crskhr
Nov 25 '18 at 14:34
$begingroup$
@Blue Thanks a lot for clarifying :)
$endgroup$
– crskhr
Nov 25 '18 at 14:34
|
show 1 more comment
$begingroup$
Because there are so many squares, coordinates are easy to compute.
The area of the shaded square is clearly $u^2+v^2$.
The area of the shaded triangle is one-half of the absolute value of the determinant of the array
$$left[ begin{array}{c}
1 & 1 & 1 \
2u-v & 3u & 2u \
3u+v & u+3v & u+v
end{array} right]$$
which is also $u^2+v^2$.
I have a second solution.
$triangle GPN$ is obtained by rotating $triangle GSD 90^circ$ clockwise.
$triangle GQM$ is obtained by rotating $triangle GRK 90^circ$ counterclockwise.
answered Nov 23 '18 at 14:50
steven gregorysteven gregory
18k32258
$endgroup$
add a comment |
$begingroup$
Because there are so many squares, coordinates are easy to compute.
The area of the shaded square is clearly $u^2+v^2$.
The area of the shaded triangle is one-half of the absolute value of the determinant of the array
$$left[ begin{array}{c}
1 & 1 & 1 \
2u-v & 3u & 2u \
3u+v & u+3v & u+v
end{array} right]$$
which is also $u^2+v^2$.
I have a second solution.
$triangle GPN$ is obtained by rotating $triangle GSD 90^circ$ clockwise.
$triangle GQM$ is obtained by rotating $triangle GRK 90^circ$ counterclockwise.
answered Nov 23 '18 at 14:50
steven gregorysteven gregory
18k32258
$endgroup$
add a comment |
$begingroup$
Because there are so many squares, coordinates are easy to compute.
The area of the shaded square is clearly $u^2+v^2$.
The area of the shaded triangle is one-half of the absolute value of the determinant of the array
$$left[ begin{array}{c}
1 & 1 & 1 \
2u-v & 3u & 2u \
3u+v & u+3v & u+v
end{array} right]$$
which is also $u^2+v^2$.
I have a second solution.
$triangle GPN$ is obtained by rotating $triangle GSD 90^circ$ clockwise.
$triangle GQM$ is obtained by rotating $triangle GRK 90^circ$ counterclockwise.
answered Nov 23 '18 at 14:50
steven gregorysteven gregory
18k32258
$endgroup$
Because there are so many squares, coordinates are easy to compute.
The area of the shaded square is clearly $u^2+v^2$.
The area of the shaded triangle is one-half of the absolute value of the determinant of the array
$$left[ begin{array}{c}
1 & 1 & 1 \
2u-v & 3u & 2u \
3u+v & u+3v & u+v
end{array} right]$$
which is also $u^2+v^2$.
I have a second solution.
$triangle GPN$ is obtained by rotating $triangle GSD 90^circ$ clockwise.
$triangle GQM$ is obtained by rotating $triangle GRK 90^circ$ counterclockwise.
answered Nov 23 '18 at 14:50
steven gregorysteven gregory
18k32258
edited Dec 19 '18 at 4:21
answered Nov 23 '18 at 14:50
steven gregorysteven gregory
18k32258
answered Nov 23 '18 at 14:50
steven gregorysteven gregory
18k32258
answered Nov 23 '18 at 14:50
steven gregorysteven gregory
18k32258
18k32258
add a comment |
add a comment |
$begingroup$
The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.
What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.
answered Nov 24 '18 at 9:25
Will OrrickWill Orrick
13.6k13360
$endgroup$
add a comment |
$begingroup$
The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.
What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.
answered Nov 24 '18 at 9:25
Will OrrickWill Orrick
13.6k13360
$endgroup$
add a comment |
$begingroup$
The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.
What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.
answered Nov 24 '18 at 9:25
Will OrrickWill Orrick
13.6k13360
$endgroup$
The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.
What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.
answered Nov 24 '18 at 9:25
Will OrrickWill Orrick
13.6k13360
edited Nov 24 '18 at 15:01
answered Nov 24 '18 at 9:25
Will OrrickWill Orrick
13.6k13360
answered Nov 24 '18 at 9:25
Will OrrickWill Orrick
13.6k13360
answered Nov 24 '18 at 9:25
Will OrrickWill Orrick
13.6k13360
13.6k13360
add a comment |
add a comment |
$begingroup$
While the other solutions are obviously correct, they are also unnecessarily complicated.
Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.
*) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.
While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct
answered Nov 23 '18 at 19:06
DenDenDoDenDenDo
58948
$endgroup$
7
$begingroup$
That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
$endgroup$
– Owen
Nov 23 '18 at 23:19
6
$begingroup$
Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
$endgroup$
– IanF1
Nov 24 '18 at 8:01
1
$begingroup$
I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
$endgroup$
– user3445853
Nov 24 '18 at 20:38
2
$begingroup$
If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
$endgroup$
– user3445853
Nov 24 '18 at 20:40
1
$begingroup$
The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
$endgroup$
– Blue
Dec 16 '18 at 19:53
add a comment |
$begingroup$
While the other solutions are obviously correct, they are also unnecessarily complicated.
Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.
*) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.
While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct
answered Nov 23 '18 at 19:06
DenDenDoDenDenDo
58948
$endgroup$
7
$begingroup$
That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
$endgroup$
– Owen
Nov 23 '18 at 23:19
6
$begingroup$
Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
$endgroup$
– IanF1
Nov 24 '18 at 8:01
1
$begingroup$
I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
$endgroup$
– user3445853
Nov 24 '18 at 20:38
2
$begingroup$
If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
$endgroup$
– user3445853
Nov 24 '18 at 20:40
1
$begingroup$
The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
$endgroup$
– Blue
Dec 16 '18 at 19:53
add a comment |
$begingroup$
While the other solutions are obviously correct, they are also unnecessarily complicated.
Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.
*) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.
While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct
answered Nov 23 '18 at 19:06
DenDenDoDenDenDo
58948
$endgroup$
While the other solutions are obviously correct, they are also unnecessarily complicated.
Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.
*) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.
While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct
answered Nov 23 '18 at 19:06
DenDenDoDenDenDo
58948
edited Nov 24 '18 at 9:49
answered Nov 23 '18 at 19:06
DenDenDoDenDenDo
58948
answered Nov 23 '18 at 19:06
DenDenDoDenDenDo
58948
answered Nov 23 '18 at 19:06
DenDenDoDenDenDo
58948
58948
7
$begingroup$
That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
$endgroup$
– Owen
Nov 23 '18 at 23:19
6
$begingroup$
Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
$endgroup$
– IanF1
Nov 24 '18 at 8:01
1
$begingroup$
I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
$endgroup$
– user3445853
Nov 24 '18 at 20:38
2
$begingroup$
If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
$endgroup$
– user3445853
Nov 24 '18 at 20:40
1
$begingroup$
The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
$endgroup$
– Blue
Dec 16 '18 at 19:53
add a comment |
7
$begingroup$
That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
$endgroup$
– Owen
Nov 23 '18 at 23:19
6
$begingroup$
Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
$endgroup$
– IanF1
Nov 24 '18 at 8:01
1
$begingroup$
I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
$endgroup$
– user3445853
Nov 24 '18 at 20:38
2
$begingroup$
If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
$endgroup$
– user3445853
Nov 24 '18 at 20:40
1
$begingroup$
The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
$endgroup$
– Blue
Dec 16 '18 at 19:53
7
7
$begingroup$
That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
$endgroup$
– Owen
Nov 23 '18 at 23:19
$begingroup$
That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
$endgroup$
– Owen
Nov 23 '18 at 23:19
6
6
$begingroup$
Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
$endgroup$
– IanF1
Nov 24 '18 at 8:01
$begingroup$
Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
$endgroup$
– IanF1
Nov 24 '18 at 8:01
1
1
$begingroup$
I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
$endgroup$
– user3445853
Nov 24 '18 at 20:38
$begingroup$
I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
$endgroup$
– user3445853
Nov 24 '18 at 20:38
2
2
$begingroup$
If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
$endgroup$
– user3445853
Nov 24 '18 at 20:40
$begingroup$
If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
$endgroup$
– user3445853
Nov 24 '18 at 20:40
1
1
$begingroup$
The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
$endgroup$
– Blue
Dec 16 '18 at 19:53
$begingroup$
The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
$endgroup$
– Blue
Dec 16 '18 at 19:53
add a comment |
$begingroup$
This is a long comment.
The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.
The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
answered Nov 22 '18 at 21:11
J.G.J.G.
25.6k22539
$endgroup$
$begingroup$
I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:35
$begingroup$
$BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
$endgroup$
– I like Serena
Nov 22 '18 at 21:41
$begingroup$
$BACD$ is not a rhombus.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:42
2
$begingroup$
@Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
$endgroup$
– J.G.
Nov 23 '18 at 10:22
1
$begingroup$
I have converted your answer into a comment.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:28
|
show 4 more comments
$begingroup$
This is a long comment.
The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.
The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
answered Nov 22 '18 at 21:11
J.G.J.G.
25.6k22539
$endgroup$
$begingroup$
I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:35
$begingroup$
$BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
$endgroup$
– I like Serena
Nov 22 '18 at 21:41
$begingroup$
$BACD$ is not a rhombus.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:42
2
$begingroup$
@Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
$endgroup$
– J.G.
Nov 23 '18 at 10:22
1
$begingroup$
I have converted your answer into a comment.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:28
|
show 4 more comments
$begingroup$
This is a long comment.
The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.
The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
answered Nov 22 '18 at 21:11
J.G.J.G.
25.6k22539
$endgroup$
This is a long comment.
The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.
The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
answered Nov 22 '18 at 21:11
J.G.J.G.
25.6k22539
answered Nov 22 '18 at 21:11
J.G.J.G.
25.6k22539
answered Nov 22 '18 at 21:11
J.G.J.G.
25.6k22539
answered Nov 22 '18 at 21:11
J.G.J.G.
25.6k22539
25.6k22539
$begingroup$
I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:35
$begingroup$
$BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
$endgroup$
– I like Serena
Nov 22 '18 at 21:41
$begingroup$
$BACD$ is not a rhombus.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:42
2
$begingroup$
@Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
$endgroup$
– J.G.
Nov 23 '18 at 10:22
1
$begingroup$
I have converted your answer into a comment.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:28
|
show 4 more comments
$begingroup$
I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:35
$begingroup$
$BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
$endgroup$
– I like Serena
Nov 22 '18 at 21:41
$begingroup$
$BACD$ is not a rhombus.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:42
2
$begingroup$
@Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
$endgroup$
– J.G.
Nov 23 '18 at 10:22
1
$begingroup$
I have converted your answer into a comment.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:28
$begingroup$
I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:35
$begingroup$
I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:35
$begingroup$
$BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
$endgroup$
– I like Serena
Nov 22 '18 at 21:41
$begingroup$
$BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
$endgroup$
– I like Serena
Nov 22 '18 at 21:41
$begingroup$
$BACD$ is not a rhombus.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:42
$begingroup$
$BACD$ is not a rhombus.
$endgroup$
– D. Thomine
Nov 22 '18 at 21:42
2
2
$begingroup$
@Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
$endgroup$
– J.G.
Nov 23 '18 at 10:22
$begingroup$
@Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
$endgroup$
– J.G.
Nov 23 '18 at 10:22
1
1
$begingroup$
I have converted your answer into a comment.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:28
$begingroup$
I have converted your answer into a comment.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:28
|
show 4 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009635%2fjapanese-temple-problem-from-1844%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
$endgroup$
– mickep
Nov 24 '18 at 14:03
$begingroup$
I checked it out. A great book indeed, thank you.
$endgroup$
– Larry
Nov 24 '18 at 14:22
$begingroup$
The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:24
1
$begingroup$
^ From J.G.'s answer.
$endgroup$
– TheSimpliFire
Nov 24 '18 at 14:26