Let B = (−a1,0) and C = (a2,0) with a1+ a2= a = |BC|, b = |AC| and c = |AB| and let l be the side of the square

Problem 12165

(American Mathematical Monthly, Vol.127, February 2020) Proposed by Tran Quang Hung and Nguyen Minh Ha (Vietnam).

Let M N P Q be a square with center K inscribed in triangle ABC with N and P lying on sides AB and AC, respectively, while M and Q lie on side BC. Let the incircle of BM N touch side BM at S and side BN at F , and let the incircle of CQP touch side CQ at T and side CP at E. Let L be the point of intersection of lines F S and ET . Prove that KL bisects the segment ST .

A

B C

N P

M

S Q T

L K

J F E

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. We set up a coordinate system with the x-axis along the side BC such that A = (0, h).

Let B = (−a1,0) and C = (a2,0) with a1+ a2= a = |BC|, b = |AC| and c = |AB| and let l be the side of the square. Since the triangle ABC is similar to the triangle AN P we have that ^h_a = ^h−l_l and we find that l = _a+h^ah . Then the four vertices of the square and its center are

M = l

a(−a1,0) , Q = l

a(a2,0) , N = l

a(−a1, a) , P = l

a(a2, a) , K = l

2a(a2− a1, a) . Next we find the radii of the the incircles of the right triangles BM N and CQP :

r1= la1

a1+ h + c , r2= la2

a2+ h + b. Therefore S = (xM − r1,0) and T = (xQ+ r2,0) which implies

J =S+ T

2 = l

2a



a2− a1+ a2a

a2+ h + b− a1a a1+ h + c,0

 . Moreover the lines F S and ET are respectively

y= − x− xS

tan ^∠B₂
= −

(a1+ c)(x − xS)

h , y= x− xT

tan ^∠C₂
=

(a2+ b)(x − xT) h and their intersection point is

L= − l 2a



a1− a2+ 2c − 2b,2(a2+ b)(a1+ c)

h − 2h − a

 . Finally, it is easy to verify that (1 − t)K + tL = J for

t= ah

2((a2+ b)(a1+ c) − h²)

and we are done.