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HMMT 二月 2020 · 几何 · 第 6 题

HMMT February 2020 — Geometry — Problem 6

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Let ABC be a triangle with AB = 5, BC = 6, CA = 7. Let D be a point on ray AB beyond B such that BD = 7, E be a point on ray BC beyond C such that CE = 5, and F be a point on ray CA beyond A such that AF = 6. Compute the area of the circumcircle of DEF .
解析
  1. Let ABC be a triangle with AB = 5, BC = 6, CA = 7. Let D be a point on ray AB beyond B such that BD = 7, E be a point on ray BC beyond C such that CE = 5, and F be a point on ray CA beyond A such that AF = 6. Compute the area of the circumcircle of DEF . Proposed by: James Lin 251 Answer: π 3 Solution 1: Let I be the incenter of ABC . We claim that I is the circumcenter of DEF . F A Z X I B E Y C D To prove this, let the incircle touch AB , BC , and AC at X , Y , and Z , respectively. Noting that XB = BY = 2, Y C = CZ = 4, and ZA = AX = 3, we see that XD = Y E = ZF = 9. Thus, since ◦ IX = IY = IZ = r (where r is the inradius) and ∠ IXD = ∠ IY E = ∠ IZF = 90 , we have three congruent right triangles, and so ID = IE = IF , as desired. √ 5+6+7 Let s = = 9 be the semiperimeter. By Heron’s formula, [ ABC ] = 9(9 − 5)(9 − 6)(9 − 7) = 2 √ √ [ ABC ] 2 6 6 6, so r = = . Then the area of the circumcircle of DEF is s 3 251 2 2 2 2 2 ID π = ( IX + XD ) π = ( r + s ) π = π. 3 ′ ′ ′ ′ Solution 2: Let D be a point on ray CB beyond B such that BD = 7, and similarly define E , F . ′ ′ ′ ′ Noting that DA = E A and AF = AF , we see that DE F F is cyclic by power of a point. Similarly, ′ ′ ′ ′ EF D D and F D E E are cyclic. Now, note that the radical axes for the three circles circumscribing ′ ′ ′ these quadrilaterals are the sides of ABC , which are not concurrent. Therefore, DD F F EE is cyclic. We can deduce that the circumcenter of this circle is I in two ways: either by calculating that the ′ midpoint of D E coincides with the foot from I to BC , or by noticing that the perpendicular bisector ′ of F F is AI . The area can then be calculated the same way as the previous solution. ′ F F A I C B ′ E D ′ E D Remark. The circumcircle of DEF is the Conway circle of ABC .