Problem:
Let β³ABC be a triangle with integer side lengths and the property that β B=2β A. What is the least possible perimeter of such a triangle?
Answer Choices:
A. 13
B. 14
C. 15
D. 16
E. 17
Solution:
Let a,b, and c be the lengths of the sides opposite vertices A,B, and C, respectively. Note that a<b. Applying the Law of Sines in β³ABC, together with the identities sinB=sin(2A)=2sinAcosA and
sinC=sin(Οβ3A)=sin(3A)=(sinA)(β1+4cos2A)
give
asinAβ=b2sinAcosAβ=csinA(β1+4cos2A)β
Thus cosA=2abβ and
c=a(β1+4cos2A)=a(β1+a2b2β)
which simplifies to b2=a2+ac.
In looking for the triangle with least perimeter, it can be assumed that a and c are relatively prime, because otherwise a smaller triangle can be obtained by shrinking by a factor of gcd(a,c). Then a and a+c are relatively prime as well. Because a(a+c)=b2, the numbers a and a+c must be squares, say a=r2 and a+c=s2, where 0<r<s and gcd(r,s)=1. This gives a=r2,b=rs, and c=s2βr2.
If r=1, then b=sβ₯2 and c=s2β1β₯s+1, a violation of the Triangle Inequality. If r=2, then the least perimeter will occur when s=3, with a=22=4,b=2β
3=6, and c=32β22=5. Greater values of r lead to greater perimeters. The requested minimum perimeter is 4+6+5=(C)15β.
Note: The 4-6-5 triangle has angle measures of approximately β A=41.4β,β B=82.8β, and β C=55.8β. Triangles with the property that β B=2β A have been called VUX triangles by Fitch Cheney in an article published in 1970 in The Mathematics Teacher.
The problems on this page are the property of the MAA's American Mathematics Competitions