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Difference between revisions of "2012 IMO Problems/Problem 5"

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==Solution==
 
==Solution==
  
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Let's draw a circumcircle around triangle <math>ABC</math> (= <math>\Gamma</math>), a circle with it's center as <math>A</math> and radius as <math>AC</math> (= <math>\Gamma'</math>),
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Let <math>\Gamma</math>, <math>\Gamma'</math>, <math>\Gamma''</math> be the circumcircle of triangle <math>ABC</math>, the circle with its center as <math>A</math> and radius as <math>AC</math>, and the circle with its center as <math>B</math> and radius as <math>BC</math>, Respectively.
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a circle with its center as <math>B</math> and radius as <math>BC</math> (= <math>\Gamma''</math>).
 
 
Since the center of <math>\Gamma</math> lies on <math>BC</math>, the three circles above are coaxial to line <math>CD</math>.
 
Since the center of <math>\Gamma</math> lies on <math>BC</math>, the three circles above are coaxial to line <math>CD</math>.
  
  
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Let Line <math>AX</math> and Line <math>BX</math> collide with <math>\Gamma</math> on <math>P</math> (<math>\neq A</math>) and <math>Q</math> (<math>\neq B</math>), Respectively. Also let <math>R</math> be the point where <math>AQ</math> and <math>BP</math> intersect.
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Let Line <math>AX</math> and Line <math>BX</math> collide with <math>\Gamma</math> on <math>P</math> (<math>\neq A</math>) and <math>Q</math> (<math>\neq B</math>), Respectively. Also let <math>R = AQ \cap BP</math>.
  
  
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Then, since <math>\angle AYB = \angle AZB = 90^{\circ}</math>, by ceva's theorem, the point <math>R</math> lies on the line <math>CD</math>.
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Then, since <math>\angle AYB = \angle AZB = 90^{\circ}</math>, by ceva's theorem, <math>R</math> lies on <math>CD</math>.
  
 
Since triangles <math>ABC</math> and <math>ACD</math> are similar, <math>AL^2 = AC^2 = AD \cdot AB</math>, Thus <math>\angle ALD = \angle ABL</math>.
 
Since triangles <math>ABC</math> and <math>ACD</math> are similar, <math>AL^2 = AC^2 = AD \cdot AB</math>, Thus <math>\angle ALD = \angle ABL</math>.
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Since <math>R</math> is on the line <math>CD</math>, and the line <math>CD</math> is the concentric line of <math>\Gamma'</math> and <math>\Gamma''</math>, <math>RK^2 = RL^2</math>, Thus <math>RK = RL</math>. Since <math>RM</math> is in the middle and <math>\angle ADR = \angle BKR = 90^{\circ}</math>,
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Since <math>R</math> is on <math>CD</math>, and <math>CD</math> is the concentric line of <math>\Gamma'</math> and <math>\Gamma''</math>, <math>RK^2 = RL^2</math>, Thus <math>RK = RL</math>. Since <math>RM</math> is in the middle and <math>\angle ADR = \angle BKR = 90^{\circ}</math>,
 
we can say triangles <math>RKM</math> and <math>RLM</math> are congruent. Therefore, <math>KM = LM</math>.
 
we can say triangles <math>RKM</math> and <math>RLM</math> are congruent. Therefore, <math>KM = LM</math>.
  
  
 
Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here.
 
Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here.
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~ Latex edit by Kscv
  
  

Latest revision as of 07:30, 20 November 2023

Problem

Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M=\overline{AL}\cap \overline{BK}$. Prove that $MK=ML$.

Solution

Let $\Gamma$, $\Gamma'$, $\Gamma''$ be the circumcircle of triangle $ABC$, the circle with its center as $A$ and radius as $AC$, and the circle with its center as $B$ and radius as $BC$, Respectively. Since the center of $\Gamma$ lies on $BC$, the three circles above are coaxial to line $CD$.


Let Line $AX$ and Line $BX$ collide with $\Gamma$ on $P$ ($\neq A$) and $Q$ ($\neq B$), Respectively. Also let $R = AQ \cap BP$.


Then, since $\angle AYB = \angle AZB = 90^{\circ}$, by ceva's theorem, $R$ lies on $CD$.

Since triangles $ABC$ and $ACD$ are similar, $AL^2 = AC^2 = AD \cdot AB$, Thus $\angle ALD = \angle ABL$. In the same way, $\angle BKD = \angle BAK$.

Therefore, $\angle ARD = \angle ABQ = \angle ALD$ making $(A, R, L, D)$ concyclic. In the same way, $(B, R, K, D)$ is concyclic.


So $\angle ADR = \angle ALR = 90^{\circ}$, and in the same way $\angle BKR = 90^{\circ}$. Therefore, the lines $RK$ and $RL$ are tangent to $\Gamma'$ and $\Gamma''$, respectively.


Since $R$ is on $CD$, and $CD$ is the concentric line of $\Gamma'$ and $\Gamma''$, $RK^2 = RL^2$, Thus $RK = RL$. Since $RM$ is in the middle and $\angle ADR = \angle BKR = 90^{\circ}$, we can say triangles $RKM$ and $RLM$ are congruent. Therefore, $KM = LM$.


Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here.

~ Latex edit by Kscv


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

2012 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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