HMMT 二月 2016 · 团队赛 · 第 3 题

3. [ 30 ] Let ABC be an acute triangle with incenter I and circumcenter O . Assume that ∠ OIA = 90 . Given that AI = 97 and BC = 144, compute the area of 4 ABC . Proposed by: Evan Chen Answer: 14040 We present five different solutions and outline a sixth and seventh one. In what follows, let a = BC , 1 b = CA , c = AB as usual, and denote by r and R the inradius and circumradius. Let s = ( a + b + c ). 2 In the first five solutions we will only prove that ◦ ∠ AIO = 90 = ⇒ b + c = 2 a. Let us see how this solves the problem. This lemma implies that s = 216. If we let E be the foot of I √ 2 2 on AB , then AE = s − BC = 72, consequently the inradius is r = 97 − 72 = 65. Finally, the area is sr = 216 · 65 = 14040 . First Solution. Since OI ⊥ DA , AI = DI . Now, it is a well-known fact that DI = DB = DC (this is occasionally called “Fact 5”). Then by Ptolemy’s Theorem, DB · AC + DC · AB = DA · BC = ⇒ AC + AB = 2 BC. Second Solution. As before note that I is the midpoint of AD . Let M and N be the midpoints of AB and AC , and let the reflection of M across BI be P ; thus BM = BP . Also, M I = P I , but we know M I = N I as I lies on the circumcircle of triangle AM N . Consequently, we get P I = N I ; moreover by angle chasing we have ◦ ∠ IN C = ∠ AM I = 180 − ∠ BP I = ∠ IP C. Thus triangles IN C and P IC are congruent ( CI is a bisector) so we deduce P C = N C . Thus, 1 BC = BP + P C = BM + CN = ( AB + AC ) . 2 2 Third Solution. We appeal to Euler’s Theorem, which states that IO = R ( R − 2 r ). Thus by the Pythagorean Theorem on 4 AIO (or by Power of a Point) we may write abc 2 2 2 2 2 ( s − a ) + r = AI = R − IO = 2 Rr = 2 s with the same notations as before. Thus, we derive that ( ) 2 2 abc = 2 s ( s − a ) + r = 2( s − a ) ( s ( s − a ) + ( s − b )( s − c )) ( ) 1 2 2 2 2 = ( s − a ) ( b + c ) − a + a − ( b − c ) 2 = 2 bc ( s − a ) . From this we deduce that 2 a = b + c , and we can proceed as in the previous solution. Fourth Solution. From Fact 5 again ( DB = DI = DC ), drop perpendicular from I to AB at E ; call M the midpoint of BC . Then, by AAS congruency on AIE and CDM , we immediately get that 1 CM = AE . As AE = ( AB + AC − BC ), this gives the desired conclusion. 2 Fifth Solution. This solution avoids angle-chasing and using the fact that BI and CI are angle- bisectors. Recall the perpendicularity lemma, where 2 2 2 2 W X ⊥ Y Z ⇐⇒ W Y − W Z = XY − XZ . ′ ′ Let B be on the extension of ray CA such that AB = AB . Of course, as in the proof of the angle ′ ′ ′ ′ bisector theorem, BB ‖ AI , meaning that BB ⊥ IO . Let I be the reflection of I across A ; of course, I ′ ′ ′ 2 2 ′ 2 2 is then the incenter of triangle AB C . Now, we have B I − BI = B O − BO by the perpendicularity ′ 2 2 ′ ′ 2 ′ 2 2 ′ 2 2 2 and by power of a point B O − BO = B A · B C . Moreover BI + B I = BI + BI = 2 BA + 2 AI 1 2 2 by the median formula. Subtracting, we get BI = AI + ( AB )( AB − AC ). We have a similar 2 1 2 2 2 2 expression for CI , and subtracting the two results in BI − CI = ( AB − AC ). Finally, 2 1 2 2 2 2 BI − CI = [( BC + AB − AC ) − ( BC − AB + AC ) ] 4 from which again, the result 2 BC = AB + AC follows. 2 Sixth Solution, outline. Use complex numbers, setting I = ab + bc + ca , A = − a , etc. on the unit circle (scale the picture to fit in a unit circle; we calculate scaling factor later). Set a = 1, and let u = b + c and v = bc . Write every condition in terms of u and v , and the area in terms of u and v too. There 2 130 2 should be two equations relating u and v : 2 u + v + 1 = 0 and u = v from the right angle and the 97 144 to 97 ratio, respectively. The square area can be computed in terms of u and v , because the area itself is antisymmetric so squaring it suffices. Use the first condition to homogenize (not coincidentally 2 2 2 2 2 2 the factor (1 − b )(1 − c ) = (1 + bc ) − ( b + c ) = (1 + v ) − u from the area homogenizes perfectly. . . because AB · AC = AI · AI , where I is the A -excenter, and of course the way the problem is set up A A AI = 3 AI .), and then we find the area of the scaled down version. To find the scaling factor simply A determine | b − c | by squaring it, writing in terms again of u and v , and comparing this to the value of

Seventh Solution, outline. Trigonometric solutions are also possible. One can you write everything in ◦ terms of the angles and solve the equations; for instance, the ∠ AIO = 90 condition can be rewritten B C 2 sin 2 sin 1 B − C B C 97 2 2 as cos = 2 sin sin and the 97 to 144 ratio condition can be rewritten as = . 2 2 2 2 sin A 144 A B C The first equation implies sin = 2 sin sin , which we can plug into the second equation to get 2 2 2 A cos . 2