EGMO Problem 3.20 (BAMO 2013/3)
$begingroup$
The problem statement follows:
Let $H$ be the orthocenter of an acute angle triangle $ABC$. Consider the circumcenters of triangles $ABH$, $ BCH$, and $CAH$. Prove that they are the vertices of a triangle that is congruent to $ABC$.
So I first showed that the orthocenter of $ABC$ is the circumcenter of the second triangle, $A'B'C'$ and the circumcenter of $ABC$ is the orthocenter of $A'B'C'$. Next if we take a homothety $h$ at $N_9$ of $ABC$ with scale factor $-1$, this will send $H$ to $O$ and vice versa and we'll get a congruent triangle.
But my question is, how do I prove that the triangle formed by taking the homothety is the very triangle in the question, namely $A'B'C'$?
geometry contest-math euclidean-geometry triangle geometric-transformation
edited Feb 8 at 15:37
greedoid
42.7k1153105
asked Feb 8 at 10:31
Shafin AhmedShafin Ahmed
707
$endgroup$
add a comment |
$begingroup$
The problem statement follows:
Let $H$ be the orthocenter of an acute angle triangle $ABC$. Consider the circumcenters of triangles $ABH$, $ BCH$, and $CAH$. Prove that they are the vertices of a triangle that is congruent to $ABC$.
So I first showed that the orthocenter of $ABC$ is the circumcenter of the second triangle, $A'B'C'$ and the circumcenter of $ABC$ is the orthocenter of $A'B'C'$. Next if we take a homothety $h$ at $N_9$ of $ABC$ with scale factor $-1$, this will send $H$ to $O$ and vice versa and we'll get a congruent triangle.
But my question is, how do I prove that the triangle formed by taking the homothety is the very triangle in the question, namely $A'B'C'$?
geometry contest-math euclidean-geometry triangle geometric-transformation
edited Feb 8 at 15:37
greedoid
42.7k1153105
asked Feb 8 at 10:31
Shafin AhmedShafin Ahmed
707
$endgroup$
add a comment |
$begingroup$
The problem statement follows:
Let $H$ be the orthocenter of an acute angle triangle $ABC$. Consider the circumcenters of triangles $ABH$, $ BCH$, and $CAH$. Prove that they are the vertices of a triangle that is congruent to $ABC$.
So I first showed that the orthocenter of $ABC$ is the circumcenter of the second triangle, $A'B'C'$ and the circumcenter of $ABC$ is the orthocenter of $A'B'C'$. Next if we take a homothety $h$ at $N_9$ of $ABC$ with scale factor $-1$, this will send $H$ to $O$ and vice versa and we'll get a congruent triangle.
But my question is, how do I prove that the triangle formed by taking the homothety is the very triangle in the question, namely $A'B'C'$?
geometry contest-math euclidean-geometry triangle geometric-transformation
edited Feb 8 at 15:37
greedoid
42.7k1153105
asked Feb 8 at 10:31
Shafin AhmedShafin Ahmed
707
$endgroup$
The problem statement follows:
Let $H$ be the orthocenter of an acute angle triangle $ABC$. Consider the circumcenters of triangles $ABH$, $ BCH$, and $CAH$. Prove that they are the vertices of a triangle that is congruent to $ABC$.
So I first showed that the orthocenter of $ABC$ is the circumcenter of the second triangle, $A'B'C'$ and the circumcenter of $ABC$ is the orthocenter of $A'B'C'$. Next if we take a homothety $h$ at $N_9$ of $ABC$ with scale factor $-1$, this will send $H$ to $O$ and vice versa and we'll get a congruent triangle.
But my question is, how do I prove that the triangle formed by taking the homothety is the very triangle in the question, namely $A'B'C'$?
geometry contest-math euclidean-geometry triangle geometric-transformation
geometry contest-math euclidean-geometry triangle geometric-transformation
edited Feb 8 at 15:37
greedoid
42.7k1153105
asked Feb 8 at 10:31
Shafin AhmedShafin Ahmed
707
edited Feb 8 at 15:37
greedoid
42.7k1153105
asked Feb 8 at 10:31
Shafin AhmedShafin Ahmed
707
edited Feb 8 at 15:37
greedoid
42.7k1153105
edited Feb 8 at 15:37
greedoid
42.7k1153105
edited Feb 8 at 15:37
greedoid
42.7k1153105
42.7k1153105
asked Feb 8 at 10:31
Shafin AhmedShafin Ahmed
707
asked Feb 8 at 10:31
Shafin AhmedShafin Ahmed
707
asked Feb 8 at 10:31
Shafin AhmedShafin Ahmed
707
707
add a comment |
add a comment |
1 Answer
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$begingroup$
I bealive you know the fact that circle $BCH$ is a reflection of circle $ABC$ acros $BC$. Similary is true for the other two circles.
So $$AC' = AO = CO = CA'$$
and by easy angle chase you can see $AC'||CA'$ so $A'CAC'$ is paralelogram, so $A'C' = AC$ and we are done.
answered Feb 8 at 10:45
greedoidgreedoid
42.7k1153105
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I bealive you know the fact that circle $BCH$ is a reflection of circle $ABC$ acros $BC$. Similary is true for the other two circles.
So $$AC' = AO = CO = CA'$$
and by easy angle chase you can see $AC'||CA'$ so $A'CAC'$ is paralelogram, so $A'C' = AC$ and we are done.
answered Feb 8 at 10:45
greedoidgreedoid
42.7k1153105
$endgroup$
add a comment |
$begingroup$
I bealive you know the fact that circle $BCH$ is a reflection of circle $ABC$ acros $BC$. Similary is true for the other two circles.
So $$AC' = AO = CO = CA'$$
and by easy angle chase you can see $AC'||CA'$ so $A'CAC'$ is paralelogram, so $A'C' = AC$ and we are done.
answered Feb 8 at 10:45
greedoidgreedoid
42.7k1153105
$endgroup$
add a comment |
$begingroup$
I bealive you know the fact that circle $BCH$ is a reflection of circle $ABC$ acros $BC$. Similary is true for the other two circles.
So $$AC' = AO = CO = CA'$$
and by easy angle chase you can see $AC'||CA'$ so $A'CAC'$ is paralelogram, so $A'C' = AC$ and we are done.
answered Feb 8 at 10:45
greedoidgreedoid
42.7k1153105
$endgroup$
I bealive you know the fact that circle $BCH$ is a reflection of circle $ABC$ acros $BC$. Similary is true for the other two circles.
So $$AC' = AO = CO = CA'$$
and by easy angle chase you can see $AC'||CA'$ so $A'CAC'$ is paralelogram, so $A'C' = AC$ and we are done.
answered Feb 8 at 10:45
greedoidgreedoid
42.7k1153105
answered Feb 8 at 10:45
greedoidgreedoid
42.7k1153105
answered Feb 8 at 10:45
greedoidgreedoid
42.7k1153105
answered Feb 8 at 10:45
greedoidgreedoid
42.7k1153105
42.7k1153105
add a comment |
add a comment |
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