2017 AIME I Problems/Problem 15

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 $m+n+p.$

$[asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]$

Solution 1

Let's start by proving a 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 the two vertices of the equilateral triangle on the legs of the right 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 $75+3+67=\boxed{145}$.

Solution 2 (No Coordinates)

Let $S$ be the triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37}$.

We will think about this problem backwards, by constructing a triangle as large as possible (We will call it $T$, for convenience) which is similar to $S$ with vertices outside of a unit equilateral triangle $\triangle ABC$, such that each vertex of the equilateral triangle lies on a side of $T$. After we find the side lengths of $T$, we will use ratios to trace back towards the original problem.

First of all, let $\theta = 90^{\circ}$, $\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)$, and $\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)$ (These three angles are simply the angles of triangle $S$; out of these three angles, $\alpha$ is the smallest angle, and $\theta$ is the largest angle). Then let us consider a point $P$ inside $\triangle ABC$ such that $\angle APB = 180^{\circ} - \theta$, $\angle BPC = 180^{\circ} - \alpha$, and $\angle APC = 180^{\circ} - \beta$. Construct the circumcircles $\omega_{AB}, ~\omega_{BC},$ and $\omega_{AC}$ of triangles $APB, ~BPC,$ and $APC$ respectively.

From here, we will prove the lemma that if we choose points $X$, $Y$, and $Z$ on circumcircles $\omega_{AB}, ~\omega_{BC},$ and $\omega_{AC}$ respectively such that $X$, $B$, and $Y$ are collinear and $Y$, $C$, and $Z$ are collinear, then $Z$, $A$, and $X$ must be collinear. First of all, if we let $\angle PAX = m$, then $\angle PBX = 180^{\circ} - m$ (by the properties of cyclic quadrilaterals), $\angle PBY = m$ (by adjacent angles), $\angle PCY = 180^{\circ} - m$ (by cyclic quadrilaterals), $\angle PCZ = m$ (adjacent angles), and $\angle PAZ = 180^{\circ} - m$ (cyclic quadrilaterals). Since $\angle PAX$ and $\angle PAZ$ are supplementary, $Z$, $A$, and $X$ are collinear as desired. Hence, $\triangle XYZ$ has an inscribed equilateral triangle $ABC$.

In addition, now we know that all triangles $XYZ$ (as described above) must be similar to triangle $S$, as $\angle AXB = \theta$ and $\angle BYC = \alpha$, so we have developed $AA$ similarity between the two triangles. Thus, $\triangle XYZ$ is the triangle similar to $S$ which we were desiring. Our goal now is to maximize the length of $XY$, in order to maximize the area of $XYZ$, to achieve our original goal.

Note that, all triangles $PYX$ are similar to each other if $Y$, $B$, and $X$ are collinear. This is because $\angle PYB$ is constant, and $\angle PXB$ is also a constant value. Then we have $AA$ similarity between this set of triangles. To maximize $XY$, we can instead maximize $PY$, which is simply the diameter of $\omega_{BC}$. From there, we can determine that $\angle PBY = 90^{\circ}$, and with similar logic, $PA$, $PB$, and $PC$ are perpendicular to $ZX$, $XY$, and $YZ$ respectively We have found our desired largest possible triangle $T$.

All we have to do now is to calculate $YZ$, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within $S$. First of all, we will prove that $\angle ZPY = \angle ACB + \angle AXB$. By the properties of cyclic quadrilaterals, $\angle AXB = \angle PAB + \angle PBA$, which means that $\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC$. Now we will show that $\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC$. Note that, by cyclic quadrilaterals, $\angle YZP = \angle PAC$ and $\angle ZYP = \angle PBC$. Hence, $\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC$ (since $\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}$), proving the aforementioned claim. Then, since $\angle ACB = 60^{\circ}$ and $\angle AXB = \theta = 90^{\circ}$, $\angle ZPY = 150^{\circ}$.

Now we calculate $PY$ and $PZ$, which are simply the diameters of circumcircles $\omega_{BC}$ and $\omega_{AC}$, respectively. By the extended law of sines, $PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}$ and $PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}$.

We can now solve for $ZY$ with the law of cosines:

$$(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)$$

$$(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}$$

$$(ZY)^2 = \frac{37 \cdot 67}{300}$$

$$ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}$$

Now we will apply this discovery towards our original triangle $S$. Since the ratio between $ZY$ and the hypotenuse of $S$ is $\frac{\sqrt{67}}{10\sqrt{3}}$, the side length of the equilateral triangle inscribed within $S$ must be $\frac{10\sqrt{3}}{\sqrt{67}}$ (as $S$ is simply as scaled version of $XYZ$, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within $S$ is $\frac{75\sqrt{3}}{67}$, implying that the answer is $\boxed{145}$.

-Solution by TheBoomBox77

Solution 3

Let $\triangle ABC$ be the right triangle with sides $AB = x$, $AC = y$, and $BC = z$ and right angle at $A$.

Let an equilateral triangle touch $AB$, $AC$, and $BC$ at $D$, $E$, and $F$ respectively, having side lengths of $c$.

Now, call $AD$ as $a$ and $AE$ as $b$. Thus, $DB = x-a$ and $EC = y-b$.

By Law of Sines on triangles $\triangle DBF$ and $ECF$,

$BF = \frac{z(a\sqrt{3}+b)} {2y}$ and $CF = \frac{z(a+b\sqrt{3})} {2x}$.

Summing,

$BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z$.

Now substituting $AB = x = 2\sqrt{3}$, $AC = y = 5$, and $BC = \sqrt{37}$ and solving, $\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1$.

We seek to minimize $[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}$.

This is equivalent to minimizing $a^2+b^2$.

Using the lemma from solution 1, we conclude that $\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}$

Thus, $[DEF] = \frac{75\sqrt{3}}{67}$ and our final answer is $\boxed{145}$

- Awsomness2000

Solution 4

We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are $5$ and $2\sqrt{3}i$, respectively. Now let the vertex of the equilateral triangle on the real axis be $a$ and let the vertex of the equilateral triangle on the imaginary axis be $bi$. Then, the third vertex of the equilateral triangle is given by: $$(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i$$.

For this to be on the hypotenuse of the right triangle, we also have the following: $$\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}$$

Note that the area of the equilateral triangle is given by $\frac{\sqrt{3}(a^2+b^2)}{4}$, so we seek to minimize $a^2+b^2$. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above: $$1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}$$

Thus, the minimum we seek is simply $\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}$, so the desired answer is $\boxed{145}$.

Solution 5 (Alcumus)

In the complex plane, let the vertices of the triangle be $a = 5,$ $b = 2i \sqrt{3},$ and $c = 0.$ Let $e$ be one of the vertices, where $e$ is real. A point on the line passing through $a = 5$ and $b = 2i \sqrt{3}$ can be expressed in the form $$f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.$$We want the third vertex $d$ to lie on the line through $b$ and $c,$ which is the imaginary axis, so its real part is 0. Since the small triangle is equilateral, $d - e = \operatorname{cis} 60^\circ \cdot (f - e),$ or $$d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).$$Then the real part of $d$ is $$\frac{5(1 - t) - e}{2} - 3t + e = 0.$$Solving for $t$ in terms of $e,$ we find $$t = \frac{e + 5}{11}.$$Then $$f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,$$so $$f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,$$so \begin{align*} |f - e|^2 &= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\ &= \frac{268e^2 - 840e + 1200}{121}. \end{align*}This quadratic is minimized when $e = \frac{840}{2 \cdot 268} = \frac{105}{67},$ and the minimum is $\frac{300}{67},$ so the smallest area of the equilateral triangle is $$\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.$$

Solution 6

Employ the same complex bash as in Solution 4, but instead note that minimizing $x^2+y^2$ is the same as minimizing the distance from 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.

Solution 7

We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be $a$ and the point on the imaginary axis be $bi$. Then, we see that $(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).$ Now we switch back to Cartesian coordinates. The equation of the hypotenuse is $y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.$ This means that the point $\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)$ is on the line. Plugging the numbers in, we have $\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.$ Now, we note that the side length of the equilateral triangle is $a^2+b^2$ so it suffices to minimize that. By Cauchy-Schwarz, we have $(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.$ Thus, the area of the smallest triangle is $\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}$ so our desired answer is $\boxed{145}$.

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