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

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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>
 
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>
  
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==Solution==
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==Solution 1==
 
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>.
 
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>.
  
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<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>
 
and hence the answer is <math>75+3+67=\boxed{145}</math>.
 
and hence the answer is <math>75+3+67=\boxed{145}</math>.
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==Solution 2 (No Coordinates)==
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Let <math>S</math> be the triangle with side lengths <math>2\sqrt{3},~5,</math> and <math>\sqrt{37}</math>.
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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.
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First of all, let <math>\theta = 90^{\circ}</math>, <math>\alpha = \arctan{\frac{2\sqrt{3}}{5}}</math>, and <math>\beta = \arctan{\frac{5}{2\sqrt{3}}}</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.
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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>.
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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. Thus, 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.
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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>.
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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>.
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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>.
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We can now solve for <math>ZY</math> with the law of cosines:
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<cmath>(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)</cmath>
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<cmath>(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}</cmath>
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<cmath>(ZY)^2 = \frac{37 \cdot 67}{300}</cmath>
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<cmath>ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}</cmath>
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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>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}
 
{{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:42, 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 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 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 $75+3+67=\boxed{145}$.

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 $\theta = 90^{\circ}$, $\alpha = \arctan{\frac{2\sqrt{3}}{5}}$, and $\beta = \arctan{\frac{5}{2\sqrt{3}}}$ (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. Thus, 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 $\boxed{145}$.

See Also

2017 AIME I (Problems • Answer Key • Resources)
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