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

Revision as of 06:45, 14 January 2021

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

---

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 $\boxed{145}$ .

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

@@ Line 3: / Line 3: @@
 The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths <math>2\sqrt{3},~5,</math> and <math>\sqrt{37},</math> as shown, is <math>\frac{m\sqrt{p}}{n},</math> where <math>m,~n,</math> and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p.</math>
-==Solution==
+<asy>
-INSERT SOLUTION HERE
+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 <math>x,y</math> satisfy <math>px+qy=1</math>, then the minimal value of <math>\sqrt{x^2+y^2}</math> is <math>\frac{1}{\sqrt{p^2+q^2}}</math>.
+Proof: Recall that the distance between the point <math>(x_0,y_0)</math> and the line <math>px+qy+r = 0</math> is given by <math>\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}</math>. In particular, the distance between the origin and any point <math>(x,y)</math> on the line <math>px+qy=1</math> is at least <math>\frac{1}{\sqrt{p^2+q^2}}</math>.
+---
+Let the vertices of the right triangle be <math>(0,0),(5,0),(0,2\sqrt{3}),</math> and let <math>(a,0),(0,b)</math> be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is <math>\left(\frac{a+\sqrt{3}b}{2},\frac{\sqrt{3}a+b}{2}\right)</math>. This point must lie on the hypotenuse <math>\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1</math>, i.e. <math>a,b</math> must satisfy
+<cmath> \frac{a+\sqrt{3}b}{10}+\frac{\sqrt{3}a+b}{4\sqrt{3}} = 1,</cmath>
+which can be simplified to
+<cmath>\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.</cmath>
+By the lemma, the minimal value of <math>\sqrt{a^2+b^2}</math> is
+<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
+<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>
+and hence the answer is <math>75+3+67=\boxed{145}</math>.
+==Solution 2 (No Coordinates)==
+Let <math>S</math> be the triangle with side lengths <math>2\sqrt{3},~5,</math> and <math>\sqrt{37}</math>.
+We will think about this problem backwards, by constructing a triangle as large as possible (We will call it <math>T</math>, for convenience) which is similar to <math>S</math> with vertices outside of a unit equilateral triangle <math>\triangle ABC</math>, such that each vertex of the equilateral triangle lies on a side of <math>T</math>. After we find the side lengths of <math>T</math>, we will use ratios to trace back towards the original problem.
+First of all, let <math>\theta = 90^{\circ}</math>, <math>\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)</math>, and <math>\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)</math> (These three angles are simply the angles of triangle <math>S</math>; out of these three angles, <math>\alpha</math> is the smallest angle, and <math>\theta</math> is the largest angle). Then let us consider a point <math>P</math> inside <math>\triangle ABC</math> such that <math>\angle APB = 180^{\circ} - \theta</math>, <math>\angle BPC = 180^{\circ} - \alpha</math>, and <math>\angle APC = 180^{\circ} - \beta</math>. Construct the circumcircles <math>\omega_{AB}, ~\omega_{BC},</math> and <math>\omega_{AC}</math> of triangles <math>APB, ~BPC,</math> and <math>APC</math> respectively.
+From here, we will prove the lemma that if we choose points <math>X</math>, <math>Y</math>, and <math>Z</math> on circumcircles <math>\omega_{AB}, ~\omega_{BC},</math> and <math>\omega_{AC}</math> respectively such that <math>X</math>, <math>B</math>, and <math>Y</math> are collinear and <math>Y</math>, <math>C</math>, and <math>Z</math> are collinear, then <math>Z</math>, <math>A</math>, and <math>X</math> must be collinear. First of all, if we let <math>\angle PAX = m</math>, then <math>\angle PBX = 180^{\circ} - m</math> (by the properties of cyclic quadrilaterals), <math>\angle PBY = m</math> (by adjacent angles), <math>\angle PCY = 180^{\circ} - m</math> (by cyclic quadrilaterals), <math>\angle PCZ = m</math> (adjacent angles), and <math>\angle PAZ = 180^{\circ} - m</math> (cyclic quadrilaterals). Since <math>\angle PAX</math> and <math>\angle PAZ</math> are supplementary, <math>Z</math>, <math>A</math>, and <math>X</math> are collinear as desired. Hence, <math>\triangle XYZ</math> has an inscribed equilateral triangle <math>ABC</math>.
+In addition, now we know that all triangles <math>XYZ</math> (as described above) must be similar to triangle <math>S</math>, as <math>\angle AXB = \theta</math> and <math>\angle BYC = \alpha</math>, so we have developed <math>AA</math> similarity between the two triangles. Thus, <math>\triangle XYZ</math> is the triangle similar to <math>S</math> which we were desiring. Our goal now is to maximize the length of <math>XY</math>, in order to maximize the area of <math>XYZ</math>, to achieve our original goal.
+Note that, all triangles <math>PYX</math> are similar to each other if <math>Y</math>, <math>B</math>, and <math>X</math> are collinear. This is because <math>\angle PYB</math> is constant, and <math>\angle PXB</math> is also a constant value. Then we have <math>AA</math> similarity between this set of triangles. To maximize <math>XY</math>, we can instead maximize <math>PY</math>, which is simply the diameter of <math>\omega_{BC}</math>. From there, we can determine that <math>\angle PBY = 90^{\circ}</math>, and with similar logic, <math>PA</math>, <math>PB</math>, and <math>PC</math> are perpendicular to <math>ZX</math>, <math>XY</math>, and <math>YZ</math> respectively We have found our desired largest possible triangle <math>T</math>.
+All we have to do now is to calculate <math>YZ</math>, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within <math>S</math>. First of all, we will prove that <math>\angle ZPY = \angle ACB + \angle AXB</math>. By the properties of cyclic quadrilaterals, <math>\angle AXB = \angle PAB + \angle PBA</math>, which means that <math>\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC</math>. Now we will show that <math>\angle ZPY =  180^{\circ} - \angle PAC - \angle PBC</math>. Note that, by cyclic quadrilaterals, <math>\angle YZP = \angle PAC</math> and <math>\angle ZYP = \angle PBC</math>. Hence, <math>\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC</math> (since <math>\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}</math>), proving the aforementioned claim. Then, since <math>\angle ACB = 60^{\circ}</math> and <math>\angle AXB = \theta = 90^{\circ}</math>, <math>\angle ZPY = 150^{\circ}</math>.
+Now we calculate <math>PY</math> and <math>PZ</math>, which are simply the diameters of circumcircles <math>\omega_{BC}</math> and <math>\omega_{AC}</math>, respectively. By the extended law of sines, <math>PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}</math> and <math>PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}</math>.
+We can now solve for <math>ZY</math> with the law of cosines:
+<cmath>(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)</cmath>
+<cmath>(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}</cmath>
+<cmath>(ZY)^2 = \frac{37 \cdot 67}{300}</cmath>
+<cmath>ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}</cmath>
+Now we will apply this discovery towards our original triangle <math>S</math>. Since the ratio between <math>ZY</math> and the hypotenuse of <math>S</math> is <math>\frac{\sqrt{67}}{10\sqrt{3}}</math>, the side length of the equilateral triangle inscribed within <math>S</math> must be <math>\frac{10\sqrt{3}}{\sqrt{67}}</math> (as <math>S</math> is simply as scaled version of <math>XYZ</math>, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within <math>S</math> is <math>\frac{75\sqrt{3}}{67}</math>, implying that the answer is <math>\boxed{145}</math>.
+'''-Solution by TheBoomBox77'''
+== Solution 3 ==
+Let <math>\triangle ABC</math> be the right triangle with sides <math>AB = x</math>, <math>AC = y</math>, and <math>BC = z</math> and right angle at <math>A</math>.
+Let an equilateral triangle touch <math>AB</math>, <math>AC</math>, and <math>BC</math> at <math>D</math>, <math>E</math>, and <math>F</math> respectively, having side lengths of <math>c</math>.
+Now, call <math>AD</math> as <math>a</math> and <math>AE</math> as <math>b</math>. Thus, <math>DB = x-a</math> and <math>EC = y-b</math>.
+By Law of Sines on triangles <math>\triangle DBF</math> and <math>ECF</math>,
+<math>BF = \frac{z(a\sqrt{3}+b)} {2y}</math> and <math>CF = \frac{z(a+b\sqrt{3})} {2x}</math>.
+Summing,
+<math>BF+CF =  \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z</math>.
+Now substituting <math>AB = x = 2\sqrt{3}</math>, <math>AC = y = 5</math>, and <math>BC = \sqrt{37}</math> and solving,
+<math>\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1</math>.
+We seek to minimize <math>[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}</math>.
+This is equivalent to minimizing <math>a^2+b^2</math>.
+Using the lemma from solution 1, we conclude that <math>\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}</math>
+Thus, <math>[DEF] = \frac{75\sqrt{3}}{67}</math> and our final answer is <math>\boxed{145}</math>
+- 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 <math>5</math> and <math>2\sqrt{3}i</math>, respectively. Now let the vertex of the equilateral triangle on the real axis be <math>a</math> and let the vertex of the equilateral triangle on the imaginary axis be <math>bi</math>. Then, the third vertex of the equilateral triangle is given by:
+<cmath>(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</cmath>.
+For this to be on the hypotenuse of the right triangle, we also have the following:
+<cmath>\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}</cmath>
+Note that the area of the equilateral triangle is given by <math>\frac{\sqrt{3}(a^2+b^2)}{4}</math>, so we seek to minimize <math>a^2+b^2</math>. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:
+<cmath>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}</cmath>
+Thus, the minimum we seek is simply <math>\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}</math>, so the desired answer is <math>\boxed{145}</math>.
+== Solution 5 (Alcumus)==
+In the complex plane, let the vertices of the triangle be <math>a = 5,</math> <math>b = 2i \sqrt{3},</math> and <math>c = 0.</math> Let <math>e</math> be one of the vertices, where <math>e</math> is real. A point on the line passing through <math>a = 5</math> and <math>b = 2i \sqrt{3}</math> can be expressed in the form
+<cmath>f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.</cmath>We want the third vertex <math>d</math> to lie on the line through <math>b</math> and <math>c,</math> which is the imaginary axis, so its real part is 0.
+Since the small triangle is equilateral, <math>d - e = \operatorname{cis} 60^\circ \cdot (f - e),</math> or
+<cmath>d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).</cmath>Then the real part of <math>d</math> is
+<cmath>\frac{5(1 - t) - e}{2} - 3t + e = 0.</cmath>Solving for <math>t</math> in terms of <math>e,</math> we find
+<cmath>t = \frac{e + 5}{11}.</cmath>Then
+<cmath>f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,</cmath>so
+<cmath>f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,</cmath>so
+<cmath>\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*}</cmath>This quadratic is minimized when <math>e = \frac{840}{2 \cdot 268} = \frac{105}{67},</math> and the minimum is <math>\frac{300}{67},</math> so the smallest area of the equilateral triangle is
+<cmath>\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.</cmath>
+==Solution 6==
+Employ the same complex bash as in Solution 4, but instead note that minimizing <math>x^2+y^2</math> is the same as minimizing the distance from
+,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 <math>a</math> and the point on the imaginary axis be <math>bi</math>. Then, we see that <math>(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).</math> Now we switch back to Cartesian coordinates. The equation of the hypotenuse is <math>y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.</math> This means that the point <math>\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)</math> is on the line. Plugging the numbers in, we have <math>\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}.</math> Now, we note that the side length of the equilateral triangle is <math>a^2+b^2</math> so it suffices to minimize that. By Cauchy-Schwarz, we have <math>(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.</math> Thus, the area of the smallest triangle is <math>\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}</math> so our desired answer is <math>\boxed{145}</math>.
+(Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)
 ==See Also==
 {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}
 {{MAA Notice}}

2017 AIME I (Problems • Answer Key • Resources)
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

Page

Toolbox

Search