Difference between revisions of "1991 USAMO Problems/Problem 1"

Latest revision as of 07:58, 6 May 2023

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

Solution 1

$[asy] import olympiad; pair A, B, C, D, extensionAC; real angleABC; path braceBC; A = (0, 0); B = (2, 0); D = (1, .5); angleABC = atan(.5); //y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here: C = (6/11, 8/11); braceBC = brace(C, B, .1); label("$\mathsf{A}$", A, W); label("$\mathsf{B}$", B, E); label("$\mathsf{C}$", C, N); label("$\mathsf{D}$", D, S); label("$\mathsf{a}$", braceBC, NE); label("$\mathsf{b}$", A--C, NW); label("$\mathsf{c}$", A--B, S); label("$\mathsf{x}$", A--D, N); draw(A--B--C--cycle); draw(A--D); draw(anglemark(C, B, A)); draw(anglemark(B, A, D)); draw(anglemark(D, A, C)); draw(braceBC); [/asy]$ (diagram by integralarefun)

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—a 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 $\boxed{77}$ .

Alternate Solution

In $\triangle ABC$ let $\angle B = \beta, \angle A = 2\beta, \angle C = 180^{\circ} - 3\beta$ . From the law of sines, we have $\frac{a}{\sin 2\beta} = \frac{b}{\sin \beta} = \frac{c} {\sin (180^{\circ} - 3\beta)} = \frac{c}{\sin 3\beta}$ Thus the ratio $b : a : c = \sin\beta : \sin 2\beta : \sin 3\beta$ We can simplify $\frac{\sin 2\beta}{\sin\beta} = \frac{2\sin\beta\cos\beta}{\sin\beta} = 2\cos\beta$ Likewise, $\frac{\sin 3\beta}{\sin\beta} = \frac{\sin 2\beta\cos\beta + \sin\beta\cos 2\beta}{\sin\beta} = \frac{2\sin\beta\cos^2\beta + \sin\beta(\cos^2\beta - \sin^2\beta)}{\sin\beta}$ $= {2 \cos^2 \beta + \cos^2 \beta - \sin^2 \beta} = 4\cos^2 \beta - 1$ Letting $\gamma = \cos\beta$ , rewrite $b : a : c = 1 : 2\gamma : 4\gamma^2 - 1$

We find that to satisfy the conditions for an obtuse triangle, $\beta \in (0^\circ, 30^\circ)$ and therefore $\gamma \in \left(\frac{\sqrt{3}}{2}, 1\right)$ .

The rational number with minimum denominator (in order to minimize scaling to obtain integer solutions) above $\frac{\sqrt{3}}{2}$ is $\frac{7}{8}$ , which also has a denominator divisible by 2 (to take advantage of the coefficients of 2 and 4 in the ratio and further minimize scaling).

Inserting $\gamma = \frac{7}{8}$ into the ratio, we find $b : a : c = 1 : \frac{7}{4} : \frac{33}{16}$ . When scaled minimally to obtain integer side lengths, we find $b, a, c = 16, 28, 33$ and that the perimeter is $\boxed{77}$ .

(note by integralarefun: The part of the solution about finding $\gamma$ is not rigorous and would likely require further proof in an actual test.)

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1991 USAMO (Problems • Resources)
Preceded by First question	Followed by Problem 2
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All USAMO Problems and Solutions

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