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Difference between revisions of "1991 USAMO Problems/Problem 1"
(rmv trial-error, wik)
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== Problem ==
 
== Problem ==
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In triangle <math>ABC</math>, angle <math>A</math> is twice angle <math>B</math>, angle <math>C</math> is [[obtuse triangle|obtuse]], and the three side lengths <math>a, b, c</math> are integers.  Determine, with proof, the minimum possible [[perimeter]].
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In triangle <math>ABC</math>, angle <math>A</math> is twice angle <math>B</math>, angle <math>C</math> is obtuse, and the three side lengths <math>a, b, c</math> are integers.  Determine, with proof, the minimum possible perimeter.
 
  
 
==Solution==
 
==Solution==
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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,
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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,
 
 
<cmath>x=\frac{bc}{a}</cmath>
 
<cmath>x=\frac{bc}{a}</cmath>
 
However, from the angle bisector theorem, we have  
 
However, from the angle bisector theorem, we have  
 
<cmath>BD=\frac{ac}{b+c}</cmath>
 
<cmath>BD=\frac{ac}{b+c}</cmath>
 
but <math>\triangle ABD</math> is isosceles, so
 
but <math>\triangle ABD</math> is isosceles, so
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<cmath>x=BD\Longrightarrow \frac{bc}{a}=\frac{ac}{b+c}\Longrightarrow \boxed{a^2=b(b+c)}</cmath>
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<cmath>x=BD\Longrightarrow \frac{bc}{a}=\frac{ac}{b+c}\Longrightarrow a^2=b(b+c)</cmath>
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so all sets of side lengths which satisfy the conditions also meet the boxed condition. Notice that <math>\GCD(a, b, c)=1</math> or else we can form a triangle by dividing <math>a, b, c</math> by their GCD to get smaller integer side lengths. Since <math>a</math> is a square, <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. Trying different values we find that the smallest perimeter occurs when <math>(a, b, c)=(28, 16, 33)</math> and the perimeter is <math>\boxed{77}</math>.
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so all sets of side lengths which satisfy the conditions also meet the boxed condition.  
 +
 
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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>.
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 +
By the [[Law of Cosines]],
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<cmath>b^2 = a^2 + c^2 - 2ac\cos B</cmath>
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 +
Substituting <math>a^2 = b^2 + bc</math> yields <math>\cos B = \frac{b+c}{2a} = \frac{y}{2x}</math>. Since <math>\angle C > 90^{\circ}</math>, <math>0^{\circ} < \angle B < 30^{\circ} \Longrightarrow \sqrt{3} < \frac{y}{x} < 2</math>. For <math>x \le 3</math> there are no integer solutions. For <math>x = 4</math>, we have <math>y = 7</math> that works, so the side lengths are <math>(a, b, c)=(28, 16, 33)</math> and the minimal perimeter is <math>\boxed{77}</math>.
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== See also ==
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{{USAMO box|year=1991|before=First question|num-a=2}}
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[[Category:Olympiad Geometry Problems]]

Revision as of 13:40, 14 January 2008

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 $\boxed{77}$.

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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