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2012 IMO Problems/Problem 4: Difference between revisions

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First we construct three circles:the circumcircles of ABC() ,The circle() with centre A and radius AC and the circle( ) with centre B and radius BC.
Ummm. I'm only a middle school student and this is my first post, so i hope that i don't get this wrong ^_^
Note that the centre of lies on the midpoint of AB,so the three circles are co-axial with radical axis CD.
 
Let AX=Y=ABX=Z=BAYBZ=P .
Lets draw an circumcircle around triangle ABC (= circle ''a''), a circle with it's center as A and radius as AC (= circle ''b''),
so AYB=AZB=90,so X must be the orthocenter of ABP implying P lies on the radical axis.
a circle with it's center as B and radius as BC (= circle ''c'').
let us denote PWR(P) by the power of P wrt  .[which are equal]
Since the center of ''a'' lies on the line BC the three circles above are coaxial to line CD.
From similar triangles ABCACD  we get AC2=ADAB=AL2ALD=ABL
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.
similarly BKD=BAK
Then since angle AYB = angle AZB = 90, by ceva's theorem in the opposite way, the point R lies on the line CD.
now APD=ABZ=ALZ implying ADLP is cyclic.
 
so ALPL.....(1) meaning PL2=PWR(P).
Since triangles ABC and ACD are similar, AL^2 = AC^2 = AD X AB, so angle ALD = angle ABL
similarly we get BKPK.....(2) meaning PK2=PWR(P)
In the same way angle BKD = angle BAK
hence PK=PL.....(3)
So in total because angle ARD = angle ABQ = angle ALD, (A, R, L, D) is concyclic
so MKP=MLP[using (1),(2),(3)]
In the same way (B, R, K, D) is concyclic
proving MK=ML
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.
 
Thanks for reading my first post!
by 장성광

Revision as of 12:02, 23 January 2013

Ummm. I'm only a middle school student and this is my first post, so i hope that i don't get this wrong ^_^

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.

Thanks for reading my first post! by 장성광

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