HMMT 二月 2021 · 几何 · 第 5 题

HMMT February 2021 — Geometry — Problem 5

Let AEF be a triangle with EF = 20 and AE = AF = 21. Let B and D be points chosen on segments AE and AF , respectively, such that BD is parallel to EF . Point C is chosen in the interior of triangle [ ABCD ] AEF such that ABCD is cyclic. If BC = 3 and CD = 4, then the ratio of areas can be [ AEF ] a written as for relatively prime positive integers a, b . Compute 100 a + b . b

解析

Let AEF be a triangle with EF = 20 and AE = AF = 21. Let B and D be points chosen on segments AE and AF , respectively, such that BD is parallel to EF . Point C is chosen in the interior of triangle [ ABCD ] AEF such that ABCD is cyclic. If BC = 3 and CD = 4, then the ratio of areas can be [ AEF ] a written as for relatively prime positive integers a, b . Compute 100 a + b . b Proposed by: Akash Das Answer: 5300 ′ ′ ′ Solution 1: Rotate 4 ABC around A to 4 AB C , such that B is on segment AF . Note that as ′ ′ BD ‖ EF , AB = AD . From this, AB = AB = AD , and B = D . Note that ′ ∠ ADC = ∠ ABC = 180 − ∠ ADC, ′ ′ because ABCD is cyclic. Therefore, C, D, and C are collinear. Also, AC = AC , and ′ ′ ∠ CAC = ∠ DAC + ∠ C AD = ∠ DAC + ∠ CAB = ∠ EAF. ′ Thus, since AE = AF , 4 ACC ∼ 4 AEF . Now, we have ′ ′ [ ACC ] = [ ACD ] + [ ADC ] = [ ACD ] + [ ABC ] = [ ABCD ] . ′ 2 CC ′ ′ ′ But, [ ACC ] = · [ AEF ], and we know that CC = CD + DC = 4 + 3 = 7. Thus, 2 EF ′ 2 [ ABCD ] [ ACC ] 7 49 = = = . 2 [ AEF ] [ AEF ] 20 400 The answer is 100 · 49 + 400 = 5300. Solution 2: A B D C E F Since BD is parallel to EF and AE = AF , we have AB = AD . Since ABCD is cyclic, ∠ ABC + ◦ ∠ ADC = 180 . Thus we can glue 4 ABC and 4 ADC as shown in the diagram above to create a triangle that is similar to 4 AEF and has the same area as ABCD . The base of this triangle has length BC + CD = 3 + 4 = 7, so the desired ratio is 2 7 49 = . 2 20 400