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

m (I don't know what this is. However, I wrote it up more formally.)
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==Problem==
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Let <math>ABC</math> be a triangle with <math>\angle BCA=90^{\circ}</math>, and let <math>D</math> be the foot of the altitude from <math>C</math>. Let <math>X</math> be a point in the interior of the segment <math>CD</math>. Let <math>K</math> be the point on the segment <math>AX</math> such that <math>BK=BC</math>. Similarly, let <math>L</math> be the point on the segment <math>BX</math> such that <math>AL=AC</math>. Let <math>M=\overline{AL}\cap \overline{BK}</math>. Prove that <math>MK=ML</math>
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==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''),
 
Lets draw an circumcircle around triangle ABC (= circle ''a''), a circle with it's center as A and radius as AC (= circle ''b''),
 
a circle with it's center as B and radius as BC (= circle ''c'').
 
a circle with it's center as B and radius as BC (= circle ''c'').
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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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{{alternate solutions}}
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==See Also==
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{{IMO box|year=2012|num-b=4|num-a=6}}

Revision as of 01:28, 19 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

Lets draw an circumcircle around triangle ABC (= circle a), a circle with it's center as A and radius as AC (= circle b), a circle with it's center as B and radius as BC (= circle c). Since the center of a lies on the line BC the three circles above are coaxial to line CD. 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.

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.

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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