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

(New page: == Problem == 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 integer...)
 
Line 2: Line 2:
  
 
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.
 
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.
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==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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<cmath>x=\frac{bc}{a}</cmath>
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However, from the angle bisector theorem, we have
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<cmath>BD=\frac{ac}{b+c}</cmath>
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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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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>.

Revision as of 20:39, 13 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 \[x=BD\Longrightarrow \frac{bc}{a}=\frac{ac}{b+c}\Longrightarrow \boxed{a^2=b(b+c)}\] 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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