1991 USAMO Problems/Problem 1

Problem

In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse, and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter.

Solution

After drawing the triangle, also draw the angle bisector of $\angle A$ , and let it intersect $\overline{BC}$ at $D$ . Notice that $\triangle ADC\sim \triangle BAC$ , and let $AD=x$ . Now from similarity, $x=\frac{bc}{a}$ However, from the angle bisector theorem, we have $BD=\frac{ac}{b+c}$ but $\triangle ABD$ is isosceles, so $x=BD\Longrightarrow \frac{bc}{a}=\frac{ac}{b+c}\Longrightarrow \boxed{a^2=b(b+c)}$ so all sets of side lengths which satisfy the conditions also meet the boxed condition. Notice that $\GCD(a, b, c)=1$ (Error compiling LaTeX. Unknown error_msg) or else we can form a triangle by dividing $a, b, c$ by their GCD to get smaller integer side lengths. Since $a$ is a square, $b$ must also be a square because if it isn't, then $b$ must share a common factor with $b+c$ , meaning it also shares a common factor with $c$ , which means $a, b, c$ share a common factor, contradiction. Trying different values we find that the smallest perimeter occurs when $(a, b, c)=(28, 16, 33)$ and the perimeter is $\boxed{77}$ .

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1991 USAMO Problems/Problem 1

Problem

Solution