HMMT 二月 2016 · 团队赛 · 第 3 题

英文解析

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 ,
   b = CA , c = AB as usual, and denote by r and R the inradius and circumradius. Let s = ( a + b + c ).1
   In the first five solutions we will only prove that 2
   °
   ∠ 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 areais sr = 216 · 65 = 14040 .
   First Solution. Since OI ⊥ DA , AI = DI . Now, it is a well-known fact that DI = DB = DC (this isoccasionally 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 ABand 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 ; moreoverby 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,
   BC = BP + P C = BM + CN = ( AB + AC ) .1
   Third Solution. We appeal to Euler’s Theorem, which states that IO = R ( R − 2 r ).22
   Thus by the Pythagorean Theorem on 4 AIO (or by Power of a Point) we may writeabc
   2 2 2 2 2
   ( s − a ) + r = AI = R − IO = 2 Rr =
   2 swith 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 ))
   ( )
   2 2 2 21 = ( s − a ) ( b + c ) − a + a − ( b − c ) = 2 bc ( s − a ) .2
   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
   CM = AE . As AE = ( AB + AC − BC ), this gives the desired conclusion.1
   Fifth Solution. This solution avoids angle-chasing and using the fact that BI and CI are angle-2
   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
   2 21
   by the median formula. Subtracting, we get BI = AI + ( AB )( AB − AC ). We have a similar
   2 2 2 212
   expression for CI , and subtracting the two results in BI − CI = ( AB − AC ). Finally,
   2 2 2 212
   BI − CI = [( BC + AB − AC ) − ( BC − AB + AC ) ]
   from which again, the result 2 BC = AB + AC follows.4
   Sixth Solution, outline. Use complex numbers, setting I = ab + bc + ca , A = − a , etc. on the unit circle 2
   (scale the picture to fit in a unit circle; we calculate scaling factor later). Set a = 1, and let u = b + cand v = bc . Write every condition in terms of u and v , and the area in terms of u and v too. There
   1302
   should be two equations relating u and v : 2 u + v + 1 = 0 and u = v from the right angle and the 2
   144 to 97 ratio, respectively. The square area can be computed in terms of u and v , because the area 97
   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 simplydetermine | b − c | by squaring it, writing in terms again of u and v , and comparing this to the value of a

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
cos .A 2