Difference between revisions of "1991 USAMO Problems/Problem 1"
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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>. | 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>. | ||
+ | |||
+ | ==Alternate Solution== | ||
+ | In <math>\triangle ABC</math> let <math>\angle B = \beta, \angle A = 2\beta, \angle C = 180^{\circ} - 3\beta</math>. From the law of sines, we have | ||
+ | <cmath> \frac{a}{\sin 2\beta} = \frac{b}{\sin \beta} = \frac{c} {\sin (180^{\circ} - 3\beta)} = \frac{c}{\sin 3\beta}</cmath> | ||
+ | Thus the ratio | ||
+ | <cmath>b : a : c = \sin\beta : \sin 2\beta : \sin 3\beta</cmath> | ||
+ | We can simplify | ||
+ | <cmath> \frac{\sin 2\beta}{\sin\beta} = \frac{2\sin\beta\cos\beta}{\sin\beta} = 2\cos\beta </cmath> | ||
+ | Likewise, | ||
+ | <cmath> \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}</cmath> | ||
+ | <cmath> = {2 \cos^2 \beta + \cos^2 \beta - \sin^2 \beta} = 4\cos^2 \beta - 1 </cmath> | ||
+ | Letting <math>\gamma = \cos\beta</math>, rewrite | ||
+ | <cmath>b : a : c = 1 : 2\gamma : 4\gamma^2 - 1</cmath> | ||
+ | |||
+ | We find that to satisfy the conditions for an obtuse triangle, <math>\beta \in (0^\circ, 30^\circ)</math> and therefore <math>\gamma \in \left(\frac{\sqrt{3}}{2}, 1\right)</math>. | ||
+ | |||
+ | The rational number with minimum denominator (in order to minimize scaling to obtain integer solutions) above <math> \frac{\sqrt{3}}{2}} </math> is <math>\frac{7}{8}</math>, 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 <math>\gamma = \frac{7}{8}</math> into the ratio, we find <math>b : a : c = 1 : \frac{7}{4} : \frac{33}{16}</math>. When scaled minimally to obtain integer side lengths, we find | ||
+ | <cmath> b, a, c = 16, 28, 33 </cmath> | ||
+ | and that the perimeter is <math>\boxed{77}</math>. | ||
== See also == | == See also == |
Revision as of 21:53, 28 April 2014
Contents
[hide]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 or else we can form a triangle by dividing
by their greatest common divisor to get smaller integer side lengths, contradicting the perimeter minimality. Since
is squared,
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. Thus we let
, so
, and we want the minimal pair
.
By the Law of Cosines,
Substituting yields
. Since
,
. For
there are no integer solutions. For
, we have
that works, so the side lengths are
and the minimal perimeter is
.
Alternate Solution
In let
. From the law of sines, we have
Thus the ratio
We can simplify
Likewise,
Letting
, rewrite
We find that to satisfy the conditions for an obtuse triangle, and therefore
.
The rational number with minimum denominator (in order to minimize scaling to obtain integer solutions) above $\frac{\sqrt{3}}{2}}$ (Error compiling LaTeX. Unknown error_msg) is , 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 into the ratio, we find
. When scaled minimally to obtain integer side lengths, we find
and that the perimeter is
.
See also
1991 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.