Right Triangle and Circle Theorem
$begingroup$
Let $ABC$ be a triangle such that $angle BAC$ is a right angle. Suppose $D$ is a point lying on $BC$ such that $BD=1$, $DC =3$ and $angle ADB=60^{circ}$, find the length of $AC$.
I was told that there is a neat way to solve the question by considering the circumcircle of $triangle ABC$. Since $angle BAC=90^{circ}$, $BC$ must be a diameter. Let $M$ be the midpoint point $BC$, then it's the centre of the circumcircle. But then I couldn't see anything substantial, please helps,
trigonometry contest-math circles
edited Jan 6 at 14:55
the_fox
2,90231538
asked May 20 '16 at 5:57
nayr ktnnayr ktn
21715
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be a triangle such that $angle BAC$ is a right angle. Suppose $D$ is a point lying on $BC$ such that $BD=1$, $DC =3$ and $angle ADB=60^{circ}$, find the length of $AC$.
I was told that there is a neat way to solve the question by considering the circumcircle of $triangle ABC$. Since $angle BAC=90^{circ}$, $BC$ must be a diameter. Let $M$ be the midpoint point $BC$, then it's the centre of the circumcircle. But then I couldn't see anything substantial, please helps,
trigonometry contest-math circles
edited Jan 6 at 14:55
the_fox
2,90231538
asked May 20 '16 at 5:57
nayr ktnnayr ktn
21715
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be a triangle such that $angle BAC$ is a right angle. Suppose $D$ is a point lying on $BC$ such that $BD=1$, $DC =3$ and $angle ADB=60^{circ}$, find the length of $AC$.
I was told that there is a neat way to solve the question by considering the circumcircle of $triangle ABC$. Since $angle BAC=90^{circ}$, $BC$ must be a diameter. Let $M$ be the midpoint point $BC$, then it's the centre of the circumcircle. But then I couldn't see anything substantial, please helps,
trigonometry contest-math circles
edited Jan 6 at 14:55
the_fox
2,90231538
asked May 20 '16 at 5:57
nayr ktnnayr ktn
21715
$endgroup$
Let $ABC$ be a triangle such that $angle BAC$ is a right angle. Suppose $D$ is a point lying on $BC$ such that $BD=1$, $DC =3$ and $angle ADB=60^{circ}$, find the length of $AC$.
I was told that there is a neat way to solve the question by considering the circumcircle of $triangle ABC$. Since $angle BAC=90^{circ}$, $BC$ must be a diameter. Let $M$ be the midpoint point $BC$, then it's the centre of the circumcircle. But then I couldn't see anything substantial, please helps,
trigonometry contest-math circles
trigonometry contest-math circles
edited Jan 6 at 14:55
the_fox
2,90231538
asked May 20 '16 at 5:57
nayr ktnnayr ktn
21715
edited Jan 6 at 14:55
the_fox
2,90231538
asked May 20 '16 at 5:57
nayr ktnnayr ktn
21715
edited Jan 6 at 14:55
the_fox
2,90231538
edited Jan 6 at 14:55
the_fox
2,90231538
edited Jan 6 at 14:55
the_fox
2,90231538
2,90231538
asked May 20 '16 at 5:57
nayr ktnnayr ktn
21715
asked May 20 '16 at 5:57
nayr ktnnayr ktn
21715
asked May 20 '16 at 5:57
nayr ktnnayr ktn
21715
21715
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
constuct a circle of radius $2,$ centerd at $0. D = (1,0), B = (2,0), c = (-2,0)$
constuct a vector v at a 60 degree angle with tail at D.
$A = (1 + |v| cos 60, |v| sin 60)\
|A|^2 = 1 + |v| + frac 14 |v|^2 + frac34 |v|^2 = 2^2\
|v|^2 + |v| - 3 = 0$
use the binomial theorm.
$|v| = frac{-1 + sqrt{13}}{2}\
|AC| = sqrt{(3 + |v| cos 60)^2 + |v|^2 sin^2 60}\
sqrt{9 + 3|v| + |v|^2)}\
sqrt{12 + 2|v| + (-3 + |v| + |v|^2)}\
sqrt{12 + 2|v|}\
sqrt{11 + sqrt{13}}\
$
answered May 20 '16 at 7:12
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
The circumcircle provides a good approach. $|DM|=1$, $|AM|=2$, and $|MC|=2$. Because $angle ADB = 60^{circ}$, $angle ADM = 120^{circ}$. We may then find $angle DAM$ using the law of sines:
$$frac{sin{angle DAM}}{|DM|} = frac{sin{angle ADM}}{|AM|} implies sin{angle DAM} = frac12 sin{120^{circ}} = frac{sqrt{3}}{4} $$
We may therefore find $angle AMD = 180^{circ} - angle DAM - angle ADM$ and therefore $angle AMC = 180^{circ} - angle AMD = angle DAM + angle ADM$. We may now use the law of cosines to find $|AC|$:
$$begin{align}|AC|^2 &= |AM|^2+|MC|^2 - 2 |AC| |AM| cos{angle AMC}\&= 2^3 [1- cos{(120^{circ}+angle DAM)}]\&= 16 sin^2{left (60^{circ}+frac12 angle DAM right )}\ &= 4 left (sin^2{left (frac12 angle DAM right )}+3 sin^2{left (frac12 angle DAM right )} + sqrt{3} sin{angle DAM}right ) \ &= 4 left (frac{4-sqrt{13}}{8} + frac{12+3 sqrt{13}}{8} + frac34 right ) \ &= 11+sqrt{13} end{align}$$
Thus
$$|AC| = sqrt{11+sqrt{13}} $$
ADDENDUM
This is even easier when one sees that $angle ABC = 1/2 angle AMC$. Also, the previous result I had was correct but not simplified enough.
answered May 20 '16 at 6:22
Ron GordonRon Gordon
123k14156267
$endgroup$
add a comment |
$begingroup$
Let AQ be a chord perpendicular to BC cutting BC at P.
If $PD = x$, then $BP = 1 – x, QP = PA = sqrt 3 x$ and $AD = 2x$ [$triangle APD$ is special angled.]
At $P, (1 – x)(x + 3) = (sqrt 3 x)^2$.
This gives $x = dfrac {sqrt {13} - 1}{4}$, after rejecting the negative length
From $triangle APO$, $(sqrt 3 x)^2 = 2^2 – (1 + x)^2$ … (*)
From $triangle APC, AC^2 = (sqrt 3 x)^2 + (x + 3)^2$ … (#)
Combining (*) and (#), we have $AC^2 = 4x + 12$.
Result follows after putting the value of $x$ back.
answered May 20 '16 at 15:41
MickMick
12.1k31641
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
constuct a circle of radius $2,$ centerd at $0. D = (1,0), B = (2,0), c = (-2,0)$
constuct a vector v at a 60 degree angle with tail at D.
$A = (1 + |v| cos 60, |v| sin 60)\
|A|^2 = 1 + |v| + frac 14 |v|^2 + frac34 |v|^2 = 2^2\
|v|^2 + |v| - 3 = 0$
use the binomial theorm.
$|v| = frac{-1 + sqrt{13}}{2}\
|AC| = sqrt{(3 + |v| cos 60)^2 + |v|^2 sin^2 60}\
sqrt{9 + 3|v| + |v|^2)}\
sqrt{12 + 2|v| + (-3 + |v| + |v|^2)}\
sqrt{12 + 2|v|}\
sqrt{11 + sqrt{13}}\
$
answered May 20 '16 at 7:12
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
constuct a circle of radius $2,$ centerd at $0. D = (1,0), B = (2,0), c = (-2,0)$
constuct a vector v at a 60 degree angle with tail at D.
$A = (1 + |v| cos 60, |v| sin 60)\
|A|^2 = 1 + |v| + frac 14 |v|^2 + frac34 |v|^2 = 2^2\
|v|^2 + |v| - 3 = 0$
use the binomial theorm.
$|v| = frac{-1 + sqrt{13}}{2}\
|AC| = sqrt{(3 + |v| cos 60)^2 + |v|^2 sin^2 60}\
sqrt{9 + 3|v| + |v|^2)}\
sqrt{12 + 2|v| + (-3 + |v| + |v|^2)}\
sqrt{12 + 2|v|}\
sqrt{11 + sqrt{13}}\
$
answered May 20 '16 at 7:12
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
constuct a circle of radius $2,$ centerd at $0. D = (1,0), B = (2,0), c = (-2,0)$
constuct a vector v at a 60 degree angle with tail at D.
$A = (1 + |v| cos 60, |v| sin 60)\
|A|^2 = 1 + |v| + frac 14 |v|^2 + frac34 |v|^2 = 2^2\
|v|^2 + |v| - 3 = 0$
use the binomial theorm.
$|v| = frac{-1 + sqrt{13}}{2}\
|AC| = sqrt{(3 + |v| cos 60)^2 + |v|^2 sin^2 60}\
sqrt{9 + 3|v| + |v|^2)}\
sqrt{12 + 2|v| + (-3 + |v| + |v|^2)}\
sqrt{12 + 2|v|}\
sqrt{11 + sqrt{13}}\
$
answered May 20 '16 at 7:12
Doug MDoug M
45.4k31954
$endgroup$
constuct a circle of radius $2,$ centerd at $0. D = (1,0), B = (2,0), c = (-2,0)$
constuct a vector v at a 60 degree angle with tail at D.
$A = (1 + |v| cos 60, |v| sin 60)\
|A|^2 = 1 + |v| + frac 14 |v|^2 + frac34 |v|^2 = 2^2\
|v|^2 + |v| - 3 = 0$
use the binomial theorm.
$|v| = frac{-1 + sqrt{13}}{2}\
|AC| = sqrt{(3 + |v| cos 60)^2 + |v|^2 sin^2 60}\
sqrt{9 + 3|v| + |v|^2)}\
sqrt{12 + 2|v| + (-3 + |v| + |v|^2)}\
sqrt{12 + 2|v|}\
sqrt{11 + sqrt{13}}\
$
answered May 20 '16 at 7:12
Doug MDoug M
45.4k31954
answered May 20 '16 at 7:12
Doug MDoug M
45.4k31954
answered May 20 '16 at 7:12
Doug MDoug M
45.4k31954
answered May 20 '16 at 7:12
Doug MDoug M
45.4k31954
45.4k31954
add a comment |
add a comment |
$begingroup$
The circumcircle provides a good approach. $|DM|=1$, $|AM|=2$, and $|MC|=2$. Because $angle ADB = 60^{circ}$, $angle ADM = 120^{circ}$. We may then find $angle DAM$ using the law of sines:
$$frac{sin{angle DAM}}{|DM|} = frac{sin{angle ADM}}{|AM|} implies sin{angle DAM} = frac12 sin{120^{circ}} = frac{sqrt{3}}{4} $$
We may therefore find $angle AMD = 180^{circ} - angle DAM - angle ADM$ and therefore $angle AMC = 180^{circ} - angle AMD = angle DAM + angle ADM$. We may now use the law of cosines to find $|AC|$:
$$begin{align}|AC|^2 &= |AM|^2+|MC|^2 - 2 |AC| |AM| cos{angle AMC}\&= 2^3 [1- cos{(120^{circ}+angle DAM)}]\&= 16 sin^2{left (60^{circ}+frac12 angle DAM right )}\ &= 4 left (sin^2{left (frac12 angle DAM right )}+3 sin^2{left (frac12 angle DAM right )} + sqrt{3} sin{angle DAM}right ) \ &= 4 left (frac{4-sqrt{13}}{8} + frac{12+3 sqrt{13}}{8} + frac34 right ) \ &= 11+sqrt{13} end{align}$$
Thus
$$|AC| = sqrt{11+sqrt{13}} $$
ADDENDUM
This is even easier when one sees that $angle ABC = 1/2 angle AMC$. Also, the previous result I had was correct but not simplified enough.
answered May 20 '16 at 6:22
Ron GordonRon Gordon
123k14156267
$endgroup$
add a comment |
$begingroup$
The circumcircle provides a good approach. $|DM|=1$, $|AM|=2$, and $|MC|=2$. Because $angle ADB = 60^{circ}$, $angle ADM = 120^{circ}$. We may then find $angle DAM$ using the law of sines:
$$frac{sin{angle DAM}}{|DM|} = frac{sin{angle ADM}}{|AM|} implies sin{angle DAM} = frac12 sin{120^{circ}} = frac{sqrt{3}}{4} $$
We may therefore find $angle AMD = 180^{circ} - angle DAM - angle ADM$ and therefore $angle AMC = 180^{circ} - angle AMD = angle DAM + angle ADM$. We may now use the law of cosines to find $|AC|$:
$$begin{align}|AC|^2 &= |AM|^2+|MC|^2 - 2 |AC| |AM| cos{angle AMC}\&= 2^3 [1- cos{(120^{circ}+angle DAM)}]\&= 16 sin^2{left (60^{circ}+frac12 angle DAM right )}\ &= 4 left (sin^2{left (frac12 angle DAM right )}+3 sin^2{left (frac12 angle DAM right )} + sqrt{3} sin{angle DAM}right ) \ &= 4 left (frac{4-sqrt{13}}{8} + frac{12+3 sqrt{13}}{8} + frac34 right ) \ &= 11+sqrt{13} end{align}$$
Thus
$$|AC| = sqrt{11+sqrt{13}} $$
ADDENDUM
This is even easier when one sees that $angle ABC = 1/2 angle AMC$. Also, the previous result I had was correct but not simplified enough.
answered May 20 '16 at 6:22
Ron GordonRon Gordon
123k14156267
$endgroup$
add a comment |
$begingroup$
The circumcircle provides a good approach. $|DM|=1$, $|AM|=2$, and $|MC|=2$. Because $angle ADB = 60^{circ}$, $angle ADM = 120^{circ}$. We may then find $angle DAM$ using the law of sines:
$$frac{sin{angle DAM}}{|DM|} = frac{sin{angle ADM}}{|AM|} implies sin{angle DAM} = frac12 sin{120^{circ}} = frac{sqrt{3}}{4} $$
We may therefore find $angle AMD = 180^{circ} - angle DAM - angle ADM$ and therefore $angle AMC = 180^{circ} - angle AMD = angle DAM + angle ADM$. We may now use the law of cosines to find $|AC|$:
$$begin{align}|AC|^2 &= |AM|^2+|MC|^2 - 2 |AC| |AM| cos{angle AMC}\&= 2^3 [1- cos{(120^{circ}+angle DAM)}]\&= 16 sin^2{left (60^{circ}+frac12 angle DAM right )}\ &= 4 left (sin^2{left (frac12 angle DAM right )}+3 sin^2{left (frac12 angle DAM right )} + sqrt{3} sin{angle DAM}right ) \ &= 4 left (frac{4-sqrt{13}}{8} + frac{12+3 sqrt{13}}{8} + frac34 right ) \ &= 11+sqrt{13} end{align}$$
Thus
$$|AC| = sqrt{11+sqrt{13}} $$
ADDENDUM
This is even easier when one sees that $angle ABC = 1/2 angle AMC$. Also, the previous result I had was correct but not simplified enough.
answered May 20 '16 at 6:22
Ron GordonRon Gordon
123k14156267
$endgroup$
The circumcircle provides a good approach. $|DM|=1$, $|AM|=2$, and $|MC|=2$. Because $angle ADB = 60^{circ}$, $angle ADM = 120^{circ}$. We may then find $angle DAM$ using the law of sines:
$$frac{sin{angle DAM}}{|DM|} = frac{sin{angle ADM}}{|AM|} implies sin{angle DAM} = frac12 sin{120^{circ}} = frac{sqrt{3}}{4} $$
We may therefore find $angle AMD = 180^{circ} - angle DAM - angle ADM$ and therefore $angle AMC = 180^{circ} - angle AMD = angle DAM + angle ADM$. We may now use the law of cosines to find $|AC|$:
$$begin{align}|AC|^2 &= |AM|^2+|MC|^2 - 2 |AC| |AM| cos{angle AMC}\&= 2^3 [1- cos{(120^{circ}+angle DAM)}]\&= 16 sin^2{left (60^{circ}+frac12 angle DAM right )}\ &= 4 left (sin^2{left (frac12 angle DAM right )}+3 sin^2{left (frac12 angle DAM right )} + sqrt{3} sin{angle DAM}right ) \ &= 4 left (frac{4-sqrt{13}}{8} + frac{12+3 sqrt{13}}{8} + frac34 right ) \ &= 11+sqrt{13} end{align}$$
Thus
$$|AC| = sqrt{11+sqrt{13}} $$
ADDENDUM
This is even easier when one sees that $angle ABC = 1/2 angle AMC$. Also, the previous result I had was correct but not simplified enough.
answered May 20 '16 at 6:22
Ron GordonRon Gordon
123k14156267
edited May 20 '16 at 17:08
answered May 20 '16 at 6:22
Ron GordonRon Gordon
123k14156267
answered May 20 '16 at 6:22
Ron GordonRon Gordon
123k14156267
answered May 20 '16 at 6:22
Ron GordonRon Gordon
123k14156267
123k14156267
add a comment |
add a comment |
$begingroup$
Let AQ be a chord perpendicular to BC cutting BC at P.
If $PD = x$, then $BP = 1 – x, QP = PA = sqrt 3 x$ and $AD = 2x$ [$triangle APD$ is special angled.]
At $P, (1 – x)(x + 3) = (sqrt 3 x)^2$.
This gives $x = dfrac {sqrt {13} - 1}{4}$, after rejecting the negative length
From $triangle APO$, $(sqrt 3 x)^2 = 2^2 – (1 + x)^2$ … (*)
From $triangle APC, AC^2 = (sqrt 3 x)^2 + (x + 3)^2$ … (#)
Combining (*) and (#), we have $AC^2 = 4x + 12$.
Result follows after putting the value of $x$ back.
answered May 20 '16 at 15:41
MickMick
12.1k31641
$endgroup$
add a comment |
$begingroup$
Let AQ be a chord perpendicular to BC cutting BC at P.
If $PD = x$, then $BP = 1 – x, QP = PA = sqrt 3 x$ and $AD = 2x$ [$triangle APD$ is special angled.]
At $P, (1 – x)(x + 3) = (sqrt 3 x)^2$.
This gives $x = dfrac {sqrt {13} - 1}{4}$, after rejecting the negative length
From $triangle APO$, $(sqrt 3 x)^2 = 2^2 – (1 + x)^2$ … (*)
From $triangle APC, AC^2 = (sqrt 3 x)^2 + (x + 3)^2$ … (#)
Combining (*) and (#), we have $AC^2 = 4x + 12$.
Result follows after putting the value of $x$ back.
answered May 20 '16 at 15:41
MickMick
12.1k31641
$endgroup$
add a comment |
$begingroup$
Let AQ be a chord perpendicular to BC cutting BC at P.
If $PD = x$, then $BP = 1 – x, QP = PA = sqrt 3 x$ and $AD = 2x$ [$triangle APD$ is special angled.]
At $P, (1 – x)(x + 3) = (sqrt 3 x)^2$.
This gives $x = dfrac {sqrt {13} - 1}{4}$, after rejecting the negative length
From $triangle APO$, $(sqrt 3 x)^2 = 2^2 – (1 + x)^2$ … (*)
From $triangle APC, AC^2 = (sqrt 3 x)^2 + (x + 3)^2$ … (#)
Combining (*) and (#), we have $AC^2 = 4x + 12$.
Result follows after putting the value of $x$ back.
answered May 20 '16 at 15:41
MickMick
12.1k31641
$endgroup$
Let AQ be a chord perpendicular to BC cutting BC at P.
If $PD = x$, then $BP = 1 – x, QP = PA = sqrt 3 x$ and $AD = 2x$ [$triangle APD$ is special angled.]
At $P, (1 – x)(x + 3) = (sqrt 3 x)^2$.
This gives $x = dfrac {sqrt {13} - 1}{4}$, after rejecting the negative length
From $triangle APO$, $(sqrt 3 x)^2 = 2^2 – (1 + x)^2$ … (*)
From $triangle APC, AC^2 = (sqrt 3 x)^2 + (x + 3)^2$ … (#)
Combining (*) and (#), we have $AC^2 = 4x + 12$.
Result follows after putting the value of $x$ back.
answered May 20 '16 at 15:41
MickMick
12.1k31641
edited Jan 6 at 14:49
answered May 20 '16 at 15:41
MickMick
12.1k31641
answered May 20 '16 at 15:41
MickMick
12.1k31641
answered May 20 '16 at 15:41
MickMick
12.1k31641
12.1k31641
add a comment |
add a comment |
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