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Difference between revisions of "2017 AIME I Problems/Problem 15"

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By the lemma, the minimal value of <math>\sqrt{a^2+b^2}</math> is
 
By the lemma, the minimal value of <math>\sqrt{a^2+b^2}</math> is
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<cmath>\frac{1}{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2} = \frac{10\sqrt{3}}{\sqrt{67}},</cmath>
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<cmath>\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},</cmath>
 
so the minimal area of the equilateral triangle is
 
so the minimal area of the equilateral triangle is
 
<cmath> \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},</cmath>
 
<cmath> \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},</cmath>

Revision as of 11:44, 9 March 2017

Problem 15

The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p.$

Solution

Lemma. If $x,y$ satisfy $px+qy=1$, then the minimal value of $\sqrt{x^2+y^2}$ is $\frac{1}{\sqrt{p^2+q^2}}$.

Proof. Recall that the distance between the point $(x_0,y_0)$ and the line $px+qy+r = 0$ is given by $\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}$. In particular, the distance between the origin and any point $(x,y)$ on the line $px+qy=1$ is at least $\frac{1}{\sqrt{p^2+q^2}}$.

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Let the vertices of the right triangle be $(0,0),(5,0),(0,2\sqrt{3}),$ and let $(a,0),(0,b)$ be two of the vertices of the equilateral triangle. Then, the third vertex of the equilateral triangle is $\left(\frac{a+\sqrt{3}b}{2},\frac{\sqrt{3}a+b}{2}\right)$. This point must lie on the hypotenuse $\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1$, i.e. $a,b$ must satisfy \[\frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}} = 1,\] which can be simplified to \[\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.\]

By the lemma, the minimal value of $\sqrt{a^2+b^2}$ is \[\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},\] so the minimal area of the equilateral triangle is \[\frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},\] and hence the answer is $75+3+67=\boxed{145}$.

See Also

2017 AIME I (Problems • Answer Key • Resources)
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