Proove that α = η
The question is to show that α = η.
The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN.
I need help, I do not know how to solve it or how to start.
(sorry for my bad english)
trigonometry triangle
edited Dec 3 '18 at 19:50
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 3 '18 at 19:29
ElsaElsa
23
locked by Aloizio Macedo♦ Dec 6 '18 at 17:08
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
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The question is to show that α = η.
The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN.
I need help, I do not know how to solve it or how to start.
(sorry for my bad english)
trigonometry triangle
edited Dec 3 '18 at 19:50
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 3 '18 at 19:29
ElsaElsa
23
locked by Aloizio Macedo♦ Dec 6 '18 at 17:08
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
I'd really appreciate an answer :)
– Elsa
Dec 3 '18 at 20:15
Are we supposed to assume that the quadrilaterals $ABCD$ and $ADC'D'$ are squares? And the part where you state that "The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN", is it part of the task, a conjecture or something you've already proved? What have you tried so far?
– Dr. Mathva
Dec 3 '18 at 20:50
The question is that at square ABCD a point M is choosen at the side of CD an a point N on the side of BC, so that the angle MAN is 45°. ( M an N are no vertices of the square). Now the lines AM and AN subtend the circumcircle of the square in the point A and Q, or P. Show that MN and PQ are parallel
– Elsa
Dec 3 '18 at 21:28
I thought that i should show that as i asked if these two angels are equal, the lines are parallel. I added some angles to illustrate and help.
– Elsa
Dec 3 '18 at 21:33
2
This exercise is from an ongoing competition (Bundeswettbewerb Mathematik), you can see the translated exercises here: math.meta.stackexchange.com/questions/29448/… Please, do not answer the question until the fourth of March, ending of the competition...
– Dr. Mathva
Dec 4 '18 at 21:53
comments disabled on deleted / locked posts / reviews |
show 4 more comments
The question is to show that α = η.
The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN.
I need help, I do not know how to solve it or how to start.
(sorry for my bad english)
trigonometry triangle
edited Dec 3 '18 at 19:50
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 3 '18 at 19:29
ElsaElsa
23
The question is to show that α = η.
The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN.
I need help, I do not know how to solve it or how to start.
(sorry for my bad english)
trigonometry triangle
trigonometry triangle
edited Dec 3 '18 at 19:50
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 3 '18 at 19:29
ElsaElsa
23
edited Dec 3 '18 at 19:50
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 3 '18 at 19:29
ElsaElsa
23
edited Dec 3 '18 at 19:50
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Dec 3 '18 at 19:50
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Dec 3 '18 at 19:50
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Dec 3 '18 at 19:29
ElsaElsa
23
asked Dec 3 '18 at 19:29
ElsaElsa
23
asked Dec 3 '18 at 19:29
ElsaElsa
23
23
locked by Aloizio Macedo♦ Dec 6 '18 at 17:08
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
locked by Aloizio Macedo♦ Dec 6 '18 at 17:08
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
I'd really appreciate an answer :)
– Elsa
Dec 3 '18 at 20:15
Are we supposed to assume that the quadrilaterals $ABCD$ and $ADC'D'$ are squares? And the part where you state that "The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN", is it part of the task, a conjecture or something you've already proved? What have you tried so far?
– Dr. Mathva
Dec 3 '18 at 20:50
The question is that at square ABCD a point M is choosen at the side of CD an a point N on the side of BC, so that the angle MAN is 45°. ( M an N are no vertices of the square). Now the lines AM and AN subtend the circumcircle of the square in the point A and Q, or P. Show that MN and PQ are parallel
– Elsa
Dec 3 '18 at 21:28
I thought that i should show that as i asked if these two angels are equal, the lines are parallel. I added some angles to illustrate and help.
– Elsa
Dec 3 '18 at 21:33
2
This exercise is from an ongoing competition (Bundeswettbewerb Mathematik), you can see the translated exercises here: math.meta.stackexchange.com/questions/29448/… Please, do not answer the question until the fourth of March, ending of the competition...
– Dr. Mathva
Dec 4 '18 at 21:53
comments disabled on deleted / locked posts / reviews |
show 4 more comments
I'd really appreciate an answer :)
– Elsa
Dec 3 '18 at 20:15
Are we supposed to assume that the quadrilaterals $ABCD$ and $ADC'D'$ are squares? And the part where you state that "The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN", is it part of the task, a conjecture or something you've already proved? What have you tried so far?
– Dr. Mathva
Dec 3 '18 at 20:50
The question is that at square ABCD a point M is choosen at the side of CD an a point N on the side of BC, so that the angle MAN is 45°. ( M an N are no vertices of the square). Now the lines AM and AN subtend the circumcircle of the square in the point A and Q, or P. Show that MN and PQ are parallel
– Elsa
Dec 3 '18 at 21:28
I thought that i should show that as i asked if these two angels are equal, the lines are parallel. I added some angles to illustrate and help.
– Elsa
Dec 3 '18 at 21:33
2
This exercise is from an ongoing competition (Bundeswettbewerb Mathematik), you can see the translated exercises here: math.meta.stackexchange.com/questions/29448/… Please, do not answer the question until the fourth of March, ending of the competition...
– Dr. Mathva
Dec 4 '18 at 21:53
I'd really appreciate an answer :)
– Elsa
Dec 3 '18 at 20:15
I'd really appreciate an answer :)
– Elsa
Dec 3 '18 at 20:15
Are we supposed to assume that the quadrilaterals $ABCD$ and $ADC'D'$ are squares? And the part where you state that "The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN", is it part of the task, a conjecture or something you've already proved? What have you tried so far?
– Dr. Mathva
Dec 3 '18 at 20:50
Are we supposed to assume that the quadrilaterals $ABCD$ and $ADC'D'$ are squares? And the part where you state that "The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN", is it part of the task, a conjecture or something you've already proved? What have you tried so far?
– Dr. Mathva
Dec 3 '18 at 20:50
The question is that at square ABCD a point M is choosen at the side of CD an a point N on the side of BC, so that the angle MAN is 45°. ( M an N are no vertices of the square). Now the lines AM and AN subtend the circumcircle of the square in the point A and Q, or P. Show that MN and PQ are parallel
– Elsa
Dec 3 '18 at 21:28
The question is that at square ABCD a point M is choosen at the side of CD an a point N on the side of BC, so that the angle MAN is 45°. ( M an N are no vertices of the square). Now the lines AM and AN subtend the circumcircle of the square in the point A and Q, or P. Show that MN and PQ are parallel
– Elsa
Dec 3 '18 at 21:28
I thought that i should show that as i asked if these two angels are equal, the lines are parallel. I added some angles to illustrate and help.
– Elsa
Dec 3 '18 at 21:33
I thought that i should show that as i asked if these two angels are equal, the lines are parallel. I added some angles to illustrate and help.
– Elsa
Dec 3 '18 at 21:33
2
2
This exercise is from an ongoing competition (Bundeswettbewerb Mathematik), you can see the translated exercises here: math.meta.stackexchange.com/questions/29448/… Please, do not answer the question until the fourth of March, ending of the competition...
– Dr. Mathva
Dec 4 '18 at 21:53
This exercise is from an ongoing competition (Bundeswettbewerb Mathematik), you can see the translated exercises here: math.meta.stackexchange.com/questions/29448/… Please, do not answer the question until the fourth of March, ending of the competition...
– Dr. Mathva
Dec 4 '18 at 21:53
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I'd really appreciate an answer :)
– Elsa
Dec 3 '18 at 20:15
Are we supposed to assume that the quadrilaterals $ABCD$ and $ADC'D'$ are squares? And the part where you state that "The Angle QAP is the same as MAN, but AQ is not as long as AM and AP not as AN", is it part of the task, a conjecture or something you've already proved? What have you tried so far?
– Dr. Mathva
Dec 3 '18 at 20:50
The question is that at square ABCD a point M is choosen at the side of CD an a point N on the side of BC, so that the angle MAN is 45°. ( M an N are no vertices of the square). Now the lines AM and AN subtend the circumcircle of the square in the point A and Q, or P. Show that MN and PQ are parallel
– Elsa
Dec 3 '18 at 21:28
I thought that i should show that as i asked if these two angels are equal, the lines are parallel. I added some angles to illustrate and help.
– Elsa
Dec 3 '18 at 21:33
2
This exercise is from an ongoing competition (Bundeswettbewerb Mathematik), you can see the translated exercises here: math.meta.stackexchange.com/questions/29448/… Please, do not answer the question until the fourth of March, ending of the competition...
– Dr. Mathva
Dec 4 '18 at 21:53