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

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 $\frac{1}{\sqrt{p^2+q^2}}$.


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 $\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)$. This point must lie on the hypotenuse $\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1$, i.e. $a,b$ must satisfy \[\frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,\] 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

Let $AB=2\sqrt{3}, BC=5$, $D$ lies on $BC$, $F$ lies on $AB$ and $E$ lies on $AC$

Set $D$ as the origin, $BD=a,BF=b$, $F$ can be expressed as $-a+bi$ in argand plane, the distance of $CD$ is $5-a$

We know that $(-a+bi)\cdot\cos(-\frac{\pi}{3})=(-\frac{a+\sqrt{3}b}{2}+\frac{\sqrt{3}a+b}{2}i)$. We know that the slope of $AC$ is $-\frac{2\sqrt{3}}{5}$, we have that $\frac{5-a-(-\frac{a+\sqrt{3}b}{2})}{(\sqrt{3}a+b)/2}=\frac{5}{2\sqrt{3}}$, after computation, we have $11b+7\sqrt{3}a=20\sqrt{3}$

Now the rest is easy with C-S inequality, $(a^2+b^2)(147+121)\geq (7\sqrt{3}a+11a)^2, a^2+b^2\geq \frac{300}{67}$ so the smallest area is $\frac{\sqrt{3}}{4}\cdot \frac{300}{67}=\frac{75\sqrt{3}}{67}$, and the answer is $\boxed{145}$


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


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

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 (Trigonometry)

2017 AIME I 15.png

Let $ED = DF = x, AC = 5, \angle BAC = \alpha,  \angle CDE = \beta.$

Then $\tan \alpha = \frac {2\sqrt{3}}{5},$ $\angle CDF = \beta – 60^\circ,\angle AED = \beta – \alpha,$ $CD = x \cos(\beta - 60^\circ).$ By Law of Sines on triangle $\triangle ADE$ we get $\hspace{20mm} AD = x \frac {\sin (\beta – \alpha)}{\sin{\alpha}}.$ \[AD + CD = AC \implies\] \[x(\frac {\sin \beta}{\tan{\alpha}} - \cos \beta) +x (\frac {\cos\beta}{2} +\frac{\sin\beta \sqrt{3}}{2}) = 5\] \[\implies \sin \beta (\frac{ \sqrt{3}}{2}+ \frac{5}{2\sqrt{3}})- \frac{\cos\beta}{2} = \frac{5}{x} \implies\] \[\frac{5}{x} = \frac{4}{\sqrt{3}} \sin \beta - \frac{1}{2} \cos\beta \le \sqrt {\frac {16}{3} + \frac {1}{4}}=\sqrt {\frac{67}{12}}\] \[\implies x \ge 10 \sqrt{\frac {3}{67}} \implies [DEF] = \frac{x^2 \sqrt{3}}{4} = \frac{75 \sqrt{3}}{67}.\] vladimir.shelomovskii@gmail.com, vvsss

Solution 5 (Complex numbers)

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: \[(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{b}{2})i\].

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 6

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, $d - e = \cos 60^\circ \cdot (f - e),$ 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 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 $\boxed{145}$.

(Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)

Solution 8

Employ the same complex bash as in Solution 5, 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 9 (Non Analytic)

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 10

Let the right triangle's lower-left point be at $O(0,0)$. Notice the 2 other points will determine a unique equilateral triangle. Let 2 points be on the $x$-axis ($B$) and the $y$-axis ($A$) and label them $(b, 0)$ and $(0, a)$ respectively. The third point ($C$) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of $a$ and $b$.

1. Find the slope of $AB$ and take the negative reciprocal of it to find the slope of the line containing $C$. Notice the line contains the midpoint of $AB$ so we can then have an equation of the line.

2. Let $AB=x.$ For $ABC$ to be an equilateral triangle, the altitude from $C$ to $AB$ must be $\frac{x\sqrt{3}}{2}.$

We then have two equations and two variables, so we can solve for $C$'s coordinates.

We can find $C(\frac{a+b\sqrt{3}}{2}), (\frac{b+a\sqrt{3}}{2}).$ Also, note that $C$ must be on the hypotenuse of the triangle $\frac{x}{5}+\frac{y}{2\sqrt{3}}=1.$ We can plug in $x$ and $y$ as the coordinates of $C$, which simplifies to


We aim to minimize the side length of the triangle, which is $\sqrt{a^2+b^2}.$ Applying the Cauchy inequality gives us

\[(a^2+b^2)(7\sqrt{3}^2+11^2)\geq (11b+7\sqrt3a)^2 = 1200\]

From which we obtain $\sqrt{a^2+b^2} \geq \sqrt{\frac{300}{67}}.$ Thus, the area of the triangle = $\frac{75\sqrt{3}}{67}$ which leads to the answer $75+3+67=\boxed{145}.$


See Also

2017 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png