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American Mathematical Monthly
Problem 10749
by Peter Y. Woo
Biola University, La Mirada, Calif.
Problem. (Proposed by Alain Grigis, Universite Paris 13, Villetaneuse, France.) Let ABC be a triangle with ÐB = 90° and ÐA = 30°. Consider a billiard path in the triangle that begins at A, reflects successively off side BC at P, off side AC at Q, off side AB at R, off side AC at S, and then ends at B.
(a) Show that AP, QR, SB are concurrent at some point X.
(b) Show that the 3 lines cross one another at 60°.
(c) Show that AX = XP+PQ+QX = XR+RS+SX = 2 XB.
Solution. (We shall avoid trigonometrical functions
altogether). Let O be the mirror image of A in BC,
D the mirror image of B in OC, E the mirror image of C in OD, F the
mirror image of D in OE. Then the triangles ABC, OBC, ODC, ODE, OFE
are congruent, and the billard path can be reflected by these multiple
reflections into the line AF. Also ACDE is a str. line, and OCE,
ODF are equilateral triangles.
Let OYZ be the equilateral triangle with Y,Z on line AF,
such that Z is between F and Y. Then DOZE º DOYC.
Hence a" = a'. Now AEZO is a cyclic
quadrilateral because ÐOZA = ÐOEA = 60°, hence a = a"
= a'.
Now the quadrilateral OFEY is cyclic because ÐOEF =
ÐOYF. Hence ÐEYF = ÐEOF = 30°.
OACY is cyclic because ÐOYA = ÐOCA = 120°. Hence ÐCYA =
ÐCOA = 30°.
Let X be the midpoint of AY, then CX is parallel to EY,
and ÐCXY = ÐEYF = 30° = ÐCYA. Hence CX = CY.
Hence DOYC º DCXA º DOZE, proving that the
point X has the same coordinates relative to DABC as Y to DODC
and as Z to DOFE, which means that AP, QR, SB are concurrent at X.
Also from DOYC º DCXA we have OY = AX,
so that AX = XY = YZ, thus proving AX = the perimeter of DXPQ
= the perimeter of DXRS. Finally, since X is the midpoint of AY
and B is the midpoint of OA, BX = OY/2 = AX/2. QED
