Difference between revisions of "2012 IMO Problems/Problem 4"
Archmage0505 (talk | contribs) (Created page with "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. Note that the centre of lies...") |
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| − | + | Ummm. I'm only a middle school student and this is my first post, so i hope that i don't get this wrong ^_^ | |
| − | + | ||
| − | Let AX | + | 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 | ||
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
