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

m (→‎Alternate Solution)
(→‎Alternate Solution: added note)
 
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==Solution==
 
==Solution==
 
===Solution 1===
 
===Solution 1===
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<asy>
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import olympiad;
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pair A, B, C, D, extensionAC;
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real angleABC;
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path braceBC;
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A = (0, 0);
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B = (2, 0);
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D = (1, .5);
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angleABC = atan(.5);
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//y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here:
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C = (6/11, 8/11);
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braceBC = brace(C, B, .1);
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label("$\mathsf{A}$", A, W);
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label("$\mathsf{B}$", B, E);
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label("$\mathsf{C}$", C, N);
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label("$\mathsf{D}$", D, S);
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label("$\mathsf{a}$", braceBC, NE);
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label("$\mathsf{b}$", A--C, NW);
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label("$\mathsf{c}$", A--B, S);
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label("$\mathsf{x}$", A--D, N);
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draw(A--B--C--cycle);
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draw(A--D);
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draw(anglemark(C, B, A));
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draw(anglemark(B, A, D));
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draw(anglemark(D, A, C));
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draw(braceBC);
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</asy>
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(diagram by integralarefun)
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After drawing the triangle, also draw the [[angle bisector]] of <math>\angle A</math>, and let it intersect <math>\overline{BC}</math> at <math>D</math>. Notice that <math>\triangle ADC\sim \triangle BAC</math>, and let <math>AD=x</math>. Now from similarity,
 
After drawing the triangle, also draw the [[angle bisector]] of <math>\angle A</math>, and let it intersect <math>\overline{BC}</math> at <math>D</math>. Notice that <math>\triangle ADC\sim \triangle BAC</math>, and let <math>AD=x</math>. Now from similarity,
 
<cmath>x=\frac{bc}{a}</cmath>
 
<cmath>x=\frac{bc}{a}</cmath>
Line 12: Line 48:
 
so all sets of side lengths which satisfy the conditions also meet the boxed condition.  
 
so all sets of side lengths which satisfy the conditions also meet the boxed condition.  
  
−
Notice that <math>\text{gcd}(a, b, c)=1</math> or else we can form a triangle by dividing <math>a, b, c</math> by their [[greatest common divisor]] to get smaller integer side lengths, contradicting the perimeter minimality. Since <math>a</math> is squared, <math>b</math> must also be a square because if it isn't, then <math>b</math> must share a common factor with <math>b+c</math>, meaning it also shares a common factor with <math>c</math>, which means <math>a, b, c</math> share a common factor, contradiction. Thus we let <math>b = x^2, b+c = y^2</math>, so <math>a = xy</math>, and we want the minimal pair <math>(x,y)</math>.  
+
Notice that <math>\text{gcd}(a, b, c)=1</math> or else we can form a triangle by dividing <math>a, b, c</math> by their [[greatest common divisor]] to get smaller integer side lengths, contradicting the perimeter minimality. Since <math>a</math> is squared, <math>b</math> must also be a square because if it isn't, then <math>b</math> must share a common factor with <math>b+c</math>, meaning it also shares a common factor with <math>c</math>, which means <math>a, b, c</math> share a common factor&#x2014;a contradiction. Thus we let <math>b = x^2, b+c = y^2</math>, so <math>a = xy</math>, and we want the minimal pair <math>(x,y)</math>.  
  
 
By the [[Law of Cosines]],  
 
By the [[Law of Cosines]],  
Line 40: Line 76:
 
<cmath> b, a, c = 16, 28, 33 </cmath>
 
<cmath> b, a, c = 16, 28, 33 </cmath>
 
and that the perimeter is <math>\boxed{77}</math>.
 
and that the perimeter is <math>\boxed{77}</math>.
 +
 +
(note by integralarefun: The part of the solution about finding <math>\gamma</math> is not rigorous and would likely require further proof in an actual test.)
  
 
{{alternate solutions}}
 
{{alternate solutions}}

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.

See Also

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

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