Difference between revisions of "1991 USAMO Problems/Problem 1"
ZzZzZzZzZzZz (talk | contribs) (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...) |
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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. | 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== | ||
+ | |||
+ | 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> | ||
+ | However, from the angle bisector theorem, we have | ||
+ | <cmath>BD=\frac{ac}{b+c}</cmath> | ||
+ | but <math>\triangle ABD</math> is isosceles, so | ||
+ | <cmath>x=BD\Longrightarrow \frac{bc}{a}=\frac{ac}{b+c}\Longrightarrow \boxed{a^2=b(b+c)}</cmath> | ||
+ | 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 19:39, 13 January 2008
Problem
In triangle , angle
is twice angle
, angle
is obtuse, and the three side lengths
are integers. Determine, with proof, the minimum possible perimeter.
Solution
After drawing the triangle, also draw the angle bisector of , and let it intersect
at
. Notice that
, and let
. Now from similarity,
However, from the angle bisector theorem, we have
but
is isosceles, so
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
by their GCD to get smaller integer side lengths. Since
is a square,
must also be a square because if it isn't, then
must share a common factor with
, meaning it also shares a common factor with
, which means
share a common factor, contradiction. Trying different values we find that the smallest perimeter occurs when
and the perimeter is
.