Problem:
Given △ABC and a point P on one of its sides, call line ℓ the splitting line of △ABC through P if ℓ passes through P and divides △ABC into two polygons of equal perimeter. Let △ABC be a triangle where BC=219 and AB and AC are positive integers. Let M and N be the midpoints of AB and AC, respectively, and suppose that the splitting lines of △ABC through M and N intersect at 30∘. Find the perimeter of △ABC.
Solution:
Answer (459):
For any △ABC where M is the midpoint of AB, the splitting line ℓ through M is parallel to the angle bisector of ∠C. Indeed, first assume without loss of generality that BC≥AC. Then let K be the intersection of ℓ and BC, and let L be the intersection of the angle bisector of ∠C and AB. Let X be the point on line BC such that AX∥CL. Because CL bisects ∠C, it follows that ∠BCL=∠LCA, and because CL∥AX, it follows that ∠CAX=∠LCA=∠BCL=∠AXC. Thus △ACX is isosceles with AC=CX. Because ℓ is a splitting line, it follows that MB+BK=KC+CA+AM=KC+CX+AM=KX+AM, and thus BK=KX and K is the midpoint of BX. It follows that MK is the midsegment of △ABX parallel to AX and MK∥AX∥CL, as claimed.
Let a,b, and c denote the lengths of sides BC,AC, and AB, respectively. Without loss of generality, assume b≥c. The acute angle between the angle bisectors of ∠C and ∠B equals 30∘, and thus by the Exterior Angle Theorem 2∠B+∠C=30∘, implying that ∠A=120∘. By the Law of Cosines
b2+c2+bc=2192=(3⋅73)2.
Assume that neither b nor c is divisible by 3 . Then b2≡c2≡1(mod3). Hence in order for the equation above to hold, bc≡1(mod3) and thus b≡c(mod3). Let b=3m+k and c=3n+k. Then
which is not 0 modulo 9 . Because 2192 is a multiple of 9 , this is impossible, and hence one of b and c is a multiple of 3 , in which case they both are. Let p=73,b=3m, and c=3n. Then m2+n2+mn=p2. Because p is a prime, m,n, and p are pairwise relatively prime. Multiplying by 4 and completing two squares gives
4m2+4n2+4mn=3(m+n)2+(m−n)2=4p2.
Let m+n=x and m−n=y. Then x>y and
3x2=4p2−y2=(2p−y)(2p+y).
Because gcd(y,p)=1, the greatest common divisor of 2p−y and 2p+y is one of 1,2, or 4 .
If gcd(2p−y,2p+y) equals 2, then there exist relatively prime integers q and r such that 2p−y=2q and 2p+y=2r. Thus 3x2=4qr and therefore one of q and r is a perfect square while the other one is 3 times a perfect square. It follows that 4p=(2p−y)+(2p+y)=2q+2r=2s2+6t2 for some positive integers s and t, so 2p=s2+3t2. However, squares only give remainders of 0 or 1 when divided by 4 , and hence s2+3t2 cannot be congruent to 2 modulo 4 . Therefore this case is impossible.
If gcd(2p−y,2p+y) equals 1 or 4 , then one of 2p−y and 2p+y is a perfect square not divisible by 3 , while the other one is 3 times a perfect square. Because 3x2+y2=4p2 and x>y, it follows that y<p. Thus 73=p<2p−y<2p+y<3p=219. Perfect squares in that range that are not multiples of 3 are 100,121, 169, and 196. Subtracting these numbers from 4p=292 leads to the four possible values of 192,171,123, and 96 for the factor of the form 3k2. Of these, only 192=3⋅64 is 3 times a perfect square.
Hence y=46 and x=80, giving m=63,n=17,b=189, and c=51. The requested perimeter is 189+51+219=459.