USA Mathematical Talent Search Solutions to Problem 5/4/16 www.usamts.org
5/4/16. Medians AD, BE , and C F of triangle AB C meet at G as shown. shown. Six small triangl triangles, es, each each with with a verte vertex x at G, are formed. We draw the circles inscribed in triangles AF G, B DG, and C DG as shown. shown. Prove Prove that that if these these three three circles circles are all congruent, then ABC is equilateral.
A
F
E G
C
B
D
Credit This problem was contributed contributed by Professor Gregory Galperin, a long-time contributor of problems to the USAMTS. involved first showing that C GD = BGD by first showing Comments Most solutions involved that the perimeters perimeters of these these triangles are equal. Students Students took a variety ariety of approache approachess to showing AF = C D, some using a purely geometric approach, as Benjamin Dozier illustrates, some using a more trigonometric approach like that of Shotaro Makisumi, and some using a little mix of the two, like Dan Li does. Solutions edited by Richard Rusczyk
∼
Solution Solution 1 by: Benjamin Benjamin Dozier (9/NM) (9/NM) A
M
N E
K
F L
G H
C
O
I
J
D
P B
BD G as they share the altitude from G to The area of C DG equals the area of BDG BC and they have bases of equal length. C DG can be dissected into OCD OC D, ODG BD G can be dissected into P BD , P DG and BP G. The and C OG. Likewi Likewise, se, BDG
USA Mathematical Talent Search Solutions to Problem 5/4/16 www.usamts.org
area of OC D equals the area of P BD because they share a base and the respective altitudes to that base are of the same length. Likewise, the area of ODG equals the area of P DG. Thus the area of C OG equals the area of BP G. Since these two triangles have altitudes of the same length, they must have bases of the same length. Therefore C G = BG B G. BD G by Side-SideWe know C D = DB , so C DG = BDG Side-Side-Side Side congruence congruence.. Furthermore urthermore,, m∠GDC + + m∠GDB = 180 and so m∠GDC = m ∠GDB = 90 . Since median AD is also the altitude, we know that ABC is isosceles with AC = AB . Now, since O and N lie on the angle bisectors of ∠C GD and ∠AGF respectively, and Also,, N L = OI and both ∠OI G and ∠C GD = ∠AGD, we know that ∠OGI = ∠N GL. Also OGI = N GL = OGH = N GK . Now OC H = OCJ OC J by Now OCH ∠GLN are right, so ASA congruence. Likewise ODI = ODJ , N LF = N M F , and N AM = N AK . All of these triangles have an altitude of common length, the inradius, which we will call r. The area of C DG is the same as the area of GF A as the medians dissect a triangle into six smaller triangles all of the same area. Thus:
◦
∼
◦
∼ ∼ ∼ ∼ ∼
∼ ∼
1 1 1 1 1 1 (2)( r )GH + + (2)( r)J D + (2)( r)C J = = (2)( r)GK + + (2)( r )AM + + (2)( r)M F 2 2 2 2 2 2 Since GH = GK : (r )J D + (r)C J = (r)AM + r (M F ) J D + C J = AM + + M F AF = C D which implies that AB = AC = C B and thus the triangle is equilateral.
Solution Solution 2 by: Shotaro Shotaro Makisumi Makisumi (9/CA) Since the centroid divides each median into segments of proportion 1 : 2, each of the six small triangles has a base that is half of and a height a third of ABC , and so they all have the same area. We know that the radii of the incircles of AF G, BDG BD G, and C DG are equal. equal. Since Since 2A = rp, where A is the area of the triangle, p is the perimeter, and r is the radius of the incircle, are all equal, these triangles all have equal perimeter. That is,
C D + C G + DG = BD B D + BG + DG = AF + AG + F G
(1)
∼
But C D = BD , so C G = BG . By SSS SSS congru congruen ence ce,, C DG = BDG BD G. This This imp impli lies es ∠C DG = 90 . We let x = F G and θ = m ∠C GD = m ∠AGF . Then we have C G = 2x. Since C DG is a right triangle, C D = 2x sin θ and DG = 2x cos θ, and so AG = 2DG = 4x cos θ. We apply apply ◦
USA Mathematical Talent Search Solutions to Problem 5/4/16 www.usamts.org
the Law of Cosines on
AF G:
(AF )2 = (F G)2 + (AG)2 (AF )2 = x2 + (4x cos θ)2 (AF )2 = x2 + 8x2 cos2 θ AF = x 1 + 8 cos cos2 θ
− 2(F G)(AG)cos(m∠AGF ) − 2x(4x cos θ)cos θ
√
Now we can rewrite the second equality of (1) as follows:
√
2x + 2x cos θ + 2x sin θ = x 1 + 8 cos cos2 θ + 4x cos θ + x Since x = 0, we can divide through by x and simplify:
1 + 2 sin sin θ 2cos θ 4sin2 θ + 4 sin sin θ + 1 + 4 cos cos2 θ 4cos θ 8sin θ cos θ 2cos2 θ cos θ + 1 + sin θ 2sin θ cos θ (sin θ + cos θ + 1)(1 2cos θ)
−
−
−
− −
−
−
√
= 1 + 8 cos cos2 θ = 1 + 8 cos2 θ = 0 = 0
This is satisfied when sin θ + cos θ + 1 = 0 or 1 2cos θ = 0. For θ (0 , 90 ), the former has no solution, since sin θ > 0 and cos θ > 0. We solve the second equation to obtain 1 cos θ = 2 Finally, AG = 4x cos θ = 2x = C G = BG . Since Since the longer longer portions portions of the medians medians are congruent, the shorter portions are also congruent, and all six smaller triangles are congruent. This occurs only if ABC is equilateral.
−
∈
◦
◦
Q.E.D.
Solution Solution 3 by: Dan Li (10/CA) (10/CA) Lemma 5.1. The inradius, r, of a triangle with sides m, n, o and angle µ opposite the side n+o m µ of length m is r = . tan 2 2 Proof. Let the triangle be M N O, with N O = m, M N = n , M O = o, and ∠N M O = µ . Let the incenter of M N O be I . Let the points points of tangency tangency on M O, M N , N O be P , Q, R, respectively. respectively.
−
O P I R
•
M
Q
N
USA Mathematical Talent Search Solutions to Problem 5/4/16 www.usamts.org
Because I is equidistant from M O and M N (I P = I Q = r), it lies on the bisector of ∠N M O. Therefore, ∠I M Q =
∠N M O
µ
=
(2) 2 2 Because the segments from one point to two points of tangency have equal length (e.g. M P = M Q), MQ = n
= 2M Q = MQ =
− QN = n − RN = n − (m − RO) n − m + P O = n − m + (o − M P ) = n − m + o − M Q n+o−m n+o−m
(3) (4) (5)
2
Thus, r = I Q = M Q(tan ∠I M Q) =
n+o
2
− m tan µ
(6)
2
Let i = GD, j = GE , k = GF , a = F A, b = EA , c = DB . Becaus Becausee the distan distance ce from the centroid ( G) to a vertex is twice the distance from the centroid to the midpoint of the side opposite the vertex, GA = 2i, GB = 2 j , GC = 2k . By the definiti definition on of median, median, F B = a , EC = b , DC D C = c . Let α = ∠C GD = ∠AGF . Because the inradii of C GD and AGF are equal and by Lemma 5.1 Lemma 5.1,,
C G + GD
2
− DC tan ∠C GD
=
2 + − tan
=
2
k
i
2
c
α
2
+ − − tan 2 2 + − tan AG i
2k + i c = 2i + k k + a = i + c
−
GF
k
2
−a
a
FA
∠AGF
2
α
(7) (8)
2
(9) (10)
It is well-known that “all the medians together divide [a triangle] into six equal parts” (http://mathworld.w (http://mathworld.wolfram.com/T olfram.com/TriangleCentroid.h riangleCentroid.html). tml). Therefore, the areas of AGF , C GD, and BGD are equal. equal. It is similarly similarly well-k well-kno nown wn that that the product product of the semisemiperimeter and the inradius equals the area of a triangle (see (7) at http://mathworld.wolfram.com/TriangleArea.html ; a proof is given at http://mathworld.wolfram.com/Inradius.html ). Since the inradii of the three triangles
USA Mathematical Talent Search Solutions to Problem 5/4/16 www.usamts.org
are equal (since their incircles are congruent) and the areas are equal, their semiperimeters must be equal. Therefore, 2i + k + a 2k + i + c 2 j + i + c = = 2 2 2
(11)
The second and third parts of (11 (11)) result in 2k + i + c 2 j + i + c = 2 2 k = j
(12) (13)
∼
Therefore, EGC = F GB by SAS (EG = j = k = F G, C G = 2k = 2 j = BG , ∠EGC = ∠F GB ). Hence, EC = F B and
∼
b = a
(14)
The first and second parts of (11 ( 11)) yield 2i + k + a 2k + i + c = 2 2 k + c = i + a
(15)
(16)
(17) (18)
Subtracting (10 (10)) from (16 (16)) yields c
= a c = a
−a
−c
Combining (14 (14)) and (18 (18), ), a = b = c 2a = 2b = 2c AB = AC = BC
Hence,
ABC is equilateral.
