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 ha = h−ll and we find that l = a+hah . 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.