1991 USAMO Problems/Problem 1

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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 $x=BD\Longrightarrow \frac{bc}{a}=\frac{ac}{b+c}\Longrightarrow a^2=b(b+c)$ so all sets of side lengths which satisfy the conditions also meet the boxed condition.

Notice that $\text{gcd}(a, b, c)=1$ or else we can form a triangle by dividing $a, b, c$ by their greatest common divisor to get smaller integer side lengths, contradicting the perimeter minimality. Since $a$ is squared, $b$ 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. Thus we let $b = x^2, b+c = y^2$ , so $a = xy$ , and we want the minimal pair $(x,y)$ .

By the Law of Cosines, $b^2 = a^2 + c^2 - 2ac\cos B$

Substituting $a^2 = b^2 + bc$ yields $\cos B = \frac{b+c}{2a} = \frac{y}{2x}$ . Since $\angle C > 90^{\circ}$ , $0^{\circ} < \angle B < 30^{\circ} \Longrightarrow \sqrt{3} < \frac{y}{x} < 2$ . For $x \le 3$ there are no integer solutions. For $x = 4$ , we have $y = 7$ that works, so the side lengths are $(a, b, c)=(28, 16, 33)$ and the minimal perimeter is $\boxed{77}$ .

1991 USAMO (Problems)
Preceded by First question	1 • 2 • 3 • 4 • 5	Followed by Problem 2
All USAMO Problems and Solutions

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1991 USAMO Problems/Problem 1

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Problem

Solution

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