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Difference between revisions of "2012 IMO Problems/Problem 5"
(latex edit)
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==Solution==
 
==Solution==
  
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Lets draw an circumcircle around triangle ABC (= circle ''a''), a circle with it's center as A and radius as AC (= circle ''b''),
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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>),
−
a circle with it's center as B and radius as BC (= circle ''c'').
+
a circle with its center as <math>B</math> and radius as <math>BC</math> (= <math>\Gamma''</math>).
−
Since the center of ''a'' lies on the line BC the three circles above are coaxial to line CD.
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Since the center of <math>\Gamma</math> lies on <math>BC</math>, the three circles above are coaxial to line <math>CD</math>.
−
Let ) Line AX and Line BX collide with ''a'' on P (not A) and Q (not B). Also let R be the point where AQ and BP intersects.
+
 
−
Then since angle AYB = angle AZB = 90, by ceva's theorem in the opposite way, the point R lies on the line CD.
+
 
 +
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.
 +
 
 +
 
 +
Then, since <math>\angle AYB = \angle AZB = 90</math>, by ceva's theorem, the point <math>R</math> lies on the line <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>.
 +
In the same way, <math>\angle BKD = \angle BAK</math>.
 +
 
 +
Therefore, <math>\angle ARD = \angle ABQ = \angle ALD</math> making <math>(A, R, L, D)</math> concyclic.
 +
In the same way, <math>(B, R, K, D)</math> is concyclic.
 +
 
 +
 
 +
So <math>\angle ADR = \angle ALR = 90</math>, and in the same way <math>\angle BKR = 90</math>. Therefore, the lines <math>RK</math> and <math>RL</math> are tangent to <math>\Gamma'</math> and <math>\Gamma''</math>, respectively.
 +
 
 +
 
 +
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</math>,
 +
we can say triangles <math>RKM</math> and <math>RLM</math> are congruent. Therefore, <math>KM = LM</math>.
  
−
Since triangles ABC and ACD are similar, AL^2 = AC^2 = AD X AB, so angle ALD = angle ABL
 
−
In the same way angle BKD = angle BAK
 
−
So in total because angle ARD = angle ABQ = angle ALD, (A, R, L, D) is concyclic
 
−
In the same way (B, R, K, D) is concyclic
 
−
So angle ADR = angle ALR = 90, and in the same way angle BKR = 90 so the line RK and RL are tangent to each ''c'' and ''b''.
 
−
Since R is on the line CD, and the line CD is the concentric line of ''b'' and ''c'', the equation RK^2 = RL^2 is true.
 
−
Which makes the result of RK = RL. Since RM is in the middle and angle ADR = angle BKR = 90,
 
−
we can say that the triangles RKM and RLM are the same. So KM = LM.
 
  
 
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.
 +
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 03:12, 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's draw a circumcircle around triangle $ABC$ (= $\Gamma$), a circle with it's center as $A$ and radius as $AC$ (= $\Gamma'$), a circle with its center as $B$ and radius as $BC$ (= $\Gamma''$). 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 $R$ be the point where $AQ$ and $BP$ intersect.


Then, since $\angle AYB = \angle AZB = 90$, by ceva's theorem, the point $R$ lies on the line $CD$.

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


Since $R$ is on the line $CD$, and the line $CD$ is the concentric line of $\Gamma'$ and $\Gamma''$, $RK^2 = RL^2$, Thus $RK = RL$. Since $RM$ is in the middle and $\angle ADR = \angle BKR = 90$, 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.


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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