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> | ||
+ | |||
+ | <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== | ==Solution 1== | ||
− | Lemma | + | 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 | + | 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 two of the | + | 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+b\sqrt{3}}{2},\frac{a\sqrt{3}+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} | + | <cmath> \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,</cmath> |
which can be simplified to | which can be simplified to | ||
<cmath>\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.</cmath> | <cmath>\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.</cmath> | ||
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and hence the answer is <math>75+3+67=\boxed{145}</math>. | and hence the answer is <math>75+3+67=\boxed{145}</math>. | ||
− | ==Solution 2 ( | + | ==Solution 2== |
+ | |||
+ | Let <math>AB=2\sqrt{3}, BC=5</math>, <math>D</math> lies on <math>BC</math>, <math>F</math> lies on <math>AB</math> and <math>E</math> lies on <math>AC</math> | ||
+ | |||
+ | Set <math>D</math> as the origin, <math>BD=a,BF=b</math>, <math>F</math> can be expressed as <math>-a+bi</math> in argand plane, the distance of <math>CD</math> is <math>5-a</math> | ||
+ | |||
+ | We know that <math>(-a+bi)\cdot\cos(-\frac{\pi}{3})=(-\frac{a+\sqrt{3}b}{2}+\frac{\sqrt{3}a+b}{2}i)</math>. We know that the slope of <math>AC</math> is <math>-\frac{2\sqrt{3}}{5}</math>, we have that <math>\frac{5-a-(-\frac{a+\sqrt{3}b}{2})}{(\sqrt{3}a+b)/2}=\frac{5}{2\sqrt{3}}</math>, after computation, we have <math>11b+7\sqrt{3}a=20\sqrt{3}</math> | ||
+ | |||
+ | Now the rest is easy with C-S inequality, <math>(a^2+b^2)(147+121)\geq (7\sqrt{3}a+11a)^2, a^2+b^2\geq \frac{300}{67}</math> so the smallest area is <math>\frac{\sqrt{3}}{4}\cdot \frac{300}{67}=\frac{75\sqrt{3}}{67}</math>, and the answer is <math>\boxed{145}</math> | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | == 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 (Trigonometry)== | ||
+ | [[File:2017 AIME I 15.png|530px|right]] | ||
+ | Let <math>ED = DF = x, AC = 5, \angle BAC = \alpha, \angle CDE = \beta. </math> | ||
+ | |||
+ | Then <math>\cot \alpha = \frac {5}{2\sqrt{3}},\angle CDF = \beta - 60^\circ,</math> | ||
+ | <cmath>\angle AED = \beta - \alpha, CD = x \cos(\beta - 60^\circ).</cmath> | ||
+ | |||
+ | By Law of Sines on triangle <math>\triangle ADE</math> we get | ||
+ | <cmath>AD = x \frac {\sin (\beta – \alpha)}{\sin{\alpha}}.</cmath> | ||
+ | <cmath>AD + CD = AC \implies </cmath> | ||
+ | <cmath>x(\frac {\sin \beta}{\tan{\alpha}} - \cos \beta) +x (\frac {\cos\beta}{2} +\frac{\sin\beta \sqrt{3}}{2}) = AC</cmath> | ||
+ | <cmath> \implies \frac{AC}{x} = \sin \beta (\cot \alpha + \frac{\sqrt{3}}{2})- \frac{\cos\beta}{2} </cmath> | ||
+ | <cmath>a \sin \beta + b \cos \beta \le \sqrt {a^2+b^2} \implies</cmath> | ||
+ | <cmath>\frac{AC}{x} \le \sqrt {\left(\cot \alpha + \frac {\sqrt3}{2}\right)^2 + \left(\frac {1}{2}\right)^2} = \sqrt {\cot^2 \alpha + \sqrt {3} \cot \alpha + 1}</cmath> | ||
+ | The smallest area <math>[DEF]</math> is | ||
+ | <cmath>\frac {\sqrt {3}}{4} \cdot x_{min}^2 = \frac {\sqrt {3} AC^2}{4 (\cot ^2 \alpha + \sqrt {3} \cot \alpha + 1)} = \frac{75 \sqrt{3}}{67}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ===Note=== | ||
+ | <math>a \sin \beta + b \cos \beta \le \sqrt {a^2+b^2}</math> follows from Cauchy-Schwarz. | ||
+ | <cmath>a \sin \beta + b \cos \beta \le \sqrt{a^2+b^2}\sqrt{\sin^2+\cos^2}=\sqrt{a^2+b^2}</cmath> | ||
+ | ~mathboy282 | ||
+ | |||
+ | == 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 <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{b}{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 6== | ||
+ | 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 = \cos 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 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) | ||
+ | |||
+ | ==Solution 8== | ||
+ | Employ the same complex bash as in Solution 5, but instead note that minimizing <math>x^2+y^2</math> 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 <math>S</math> be the triangle with side lengths <math>2\sqrt{3},~5,</math> and <math>\sqrt{37}</math>. | 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. | 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(\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. | + | 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>. | 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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'''-Solution by TheBoomBox77''' | '''-Solution by TheBoomBox77''' | ||
+ | |||
+ | ==Solution 10== | ||
+ | Let the right triangle's lower-left point be at <math>O(0,0)</math>. Notice the 2 other points will determine a unique equilateral triangle. Let 2 points be on the <math>x</math>-axis (<math>B</math>) and the <math>y</math>-axis (<math>A</math>) and label them <math>(b, 0)</math> and <math>(0, a)</math> respectively. The third point (<math>C</math>) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of <math>a</math> and <math>b</math>. | ||
+ | |||
+ | |||
+ | 1. Find the slope of <math>AB</math> and take the negative reciprocal of it to find the slope of the line containing <math>C</math>. Notice the line contains the midpoint of <math>AB</math> so we can then have an equation of the line. | ||
+ | |||
+ | 2. Let <math>AB=x.</math> For <math>ABC</math> to be an equilateral triangle, the altitude from <math>C</math> to <math>AB</math> must be <math>\frac{x\sqrt{3}}{2}.</math> | ||
+ | |||
+ | We then have two equations and two variables, so we can solve for <math>C</math>'s coordinates. | ||
+ | |||
+ | We can find <math>C(\frac{a+b\sqrt{3}}{2}), (\frac{b+a\sqrt{3}}{2}).</math> | ||
+ | Also, note that <math>C</math> must be on the hypotenuse of the triangle <math>\frac{x}{5}+\frac{y}{2\sqrt{3}}=1.</math> We can plug in <math>x</math> and <math>y</math> as the coordinates of <math>C</math>, which simplifies to | ||
+ | |||
+ | <cmath>11b+7\sqrt{3}a=20\sqrt{3}.</cmath> | ||
+ | |||
+ | We aim to minimize the side length of the triangle, which is <math>\sqrt{a^2+b^2}.</math> Applying the Cauchy inequality gives us | ||
+ | |||
+ | <cmath>(a^2+b^2)(7\sqrt{3}^2+11^2)\geq (11b+7\sqrt3a)^2 = 1200</cmath> | ||
+ | |||
+ | From which we obtain <math>\sqrt{a^2+b^2} \geq \sqrt{\frac{300}{67}}.</math> Thus, the area of the triangle = <math>\frac{75\sqrt{3}}{67}</math> which leads to the answer <math>75+3+67=\boxed{145}.</math> | ||
+ | |||
+ | -hi_im_bob | ||
+ | |||
+ | ==Solution 11== | ||
+ | [[File:AIME2017_P15.png|430px|right]] | ||
+ | The general solution to the minimal area is as following: | ||
+ | |||
+ | <cmath>A_{min}={{\sqrt{3}m^2n^2}\over{4(m^2+n^2+\sqrt{3}mn)}},</cmath> | ||
+ | |||
+ | where <math>m</math> and <math>n</math> are the two legs of the right triangle. In this particular case <math>m=5</math> and <math>n=2\sqrt{3}</math>. When we plug in these two values, we recover the correct answer of <math>\frac{75\sqrt3}{67}</math>. | ||
+ | |||
+ | The contour of the minimal area <math>A_{min}</math> is plotted as a function of leg lengths <math>m</math> and <math>n</math>, as shown on the right hand side. | ||
+ | |||
+ | ===Note=== | ||
+ | The proof of the formula can be done in a similar fashion as Solution 4, where instead of using specific values, we define variables, <math>m</math> and <math>n</math> in this case, as per the formula. | ||
+ | |||
+ | ==Solution 12 (Geometry)== | ||
+ | [[File:2017 AIME I 15b.png|400px|right]] | ||
+ | Let <math>ED = DF = x, AC = 5, \angle BAC = \alpha.</math> | ||
+ | Then <math>\cot \alpha = \frac {5}{2\sqrt{3}}.</math> | ||
+ | |||
+ | <math>P -</math> midpoint <math>DE, M - </math> midpoint <math>FE, Q -</math> circumcenter <math>\triangle AE.</math> | ||
+ | <cmath>\angle EQM = \alpha, \angle MQE = 90^\circ - \alpha, </cmath> | ||
+ | <cmath>\angle PEQ = \angle MQE + 60^\circ = 150^\circ - \alpha.</cmath> | ||
+ | <cmath>PE = \frac {x}{2}, QE = \frac {x}{2 \sin \alpha} \implies</cmath> | ||
+ | <cmath>PQ^2 = PE^2 + QE^2 - 2 PE \cdot QE \cdot \cos(150^\circ - \alpha),</cmath> | ||
+ | <cmath>PQ^2 = \frac{x^2}{4} (\cot ^2 \alpha + \sqrt {3} \cot \alpha + 1) .</cmath> | ||
+ | Points <math>P</math> and <math>Q</math> lies on bisectors of <math>CE</math> and <math>AE,</math> so <math>PQ \ge \frac {AC}{2}.</math> | ||
+ | |||
+ | The smallest area <math>[DEF]</math> is | ||
+ | <cmath>\frac {\sqrt {3}}{4} \cdot x_{min}^2 = \frac {\sqrt {3} AC^2}{4 (\cot ^2 \alpha + \sqrt {3} \cot \alpha + 1)} = \frac{75 \sqrt{3}}{67}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Solution 13 (Kinematics+Geometry)== | ||
+ | [[File:2017 AIME I 15c.png|430px|right]] | ||
+ | Let <math>ED = DF = x, AC = 5, \angle BAC = \alpha.</math> | ||
+ | Then <math>\cot \alpha = \frac {5}{2\sqrt{3}}.</math> | ||
+ | |||
+ | Let the required triangle with minimal sides <math>DEF</math> be constructed. | ||
+ | |||
+ | Let us vary its position in an acceptable way, that is, we will perform a movement in which the vertices of the <math>\triangle DEF</math> remain on the sides of the <math>\triangle ABC.</math> Any motion of a solid plane figure can be considered as rotation around some point, the center of rotation. The speed of movement of any point is perpendicular to the segment from the center of rotation to this point. With an acceptable variation, the velocities of the vertices of the <math>\triangle DEF</math> are directed along the sides <math>\triangle ABC.</math> Consequently, there is a point <math>X</math> located at the intersection of perpendiculars to the sides of <math>\triangle ABC,</math> around which <math>\triangle DEF</math> rotates. The bases of the perpendiculars dropped from <math>X</math> to the sides of <math>\triangle ABC</math> form a regular <math>\triangle DEF.</math> | ||
+ | |||
+ | Therefore <math>X</math> is the first isodynamic point of <math>\triangle ABC.</math> | ||
+ | |||
+ | It is known that <math>\angle AXB = \angle ACB + 60^\circ = 150^\circ.</math> | ||
+ | |||
+ | The points <math>A, F, X,</math> and <math>E</math> are concyclic, <math>AX</math> is the diameter, so <math>AX = \frac {x}{\sin \alpha}.</math> Similarly, <math>BX = \frac {x}{\cos \alpha}.</math> | ||
+ | <cmath>AX^2 + BX^2 - 2 AX BX \cos 150^\circ = AB^2, AC = AB \cos \alpha \implies </cmath> | ||
+ | <cmath>AC^2 = x^2 \cdot (\cot ^2 \alpha + \sqrt {3} \cot \alpha + 1).</cmath> | ||
+ | The smallest area <math>[DEF]</math> is | ||
+ | <cmath>\frac {\sqrt {3}}{4} \cdot x_{min}^2 = \frac {\sqrt {3} AC^2}{4 (\cot ^2 \alpha + \sqrt {3} \cot \alpha + 1)} = \frac{75 \sqrt{3}}{67}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==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}} |
Latest revision as of 17:11, 6 January 2024
Contents
Problem 15
The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths and
as shown, is
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
Solution 1
Lemma: If satisfy
, then the minimal value of
is
.
Proof: Recall that the distance between the point and the line
is given by
. In particular, the distance between the origin and any point
on the line
is at least
.
---
Let the vertices of the right triangle be and let
be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is
. This point must lie on the hypotenuse
, i.e.
must satisfy
which can be simplified to
By the lemma, the minimal value of is
so the minimal area of the equilateral triangle is
and hence the answer is
.
Solution 2
Let ,
lies on
,
lies on
and
lies on
Set as the origin,
,
can be expressed as
in argand plane, the distance of
is
We know that . We know that the slope of
is
, we have that
, after computation, we have
Now the rest is easy with C-S inequality, so the smallest area is
, and the answer is
~bluesoul
Solution 3
Let be the right triangle with sides
,
, and
and right angle at
.
Let an equilateral triangle touch ,
, and
at
,
, and
respectively, having side lengths of
.
Now, call as
and
as
. Thus,
and
.
By Law of Sines on triangles and
,
and
.
Summing,
.
Now substituting ,
, and
and solving,
.
We seek to minimize .
This is equivalent to minimizing .
Using the lemma from solution 1, we conclude that
Thus, and our final answer is
- Awsomness2000
Solution 4 (Trigonometry)
Let
Then
By Law of Sines on triangle we get
The smallest area
is
vladimir.shelomovskii@gmail.com, vvsss
Note
follows from Cauchy-Schwarz.
~mathboy282
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 and
, respectively. Now let the vertex of the equilateral triangle on the real axis be
and let the vertex of the equilateral triangle on the imaginary axis be
. Then, the third vertex of the equilateral triangle is given by:
.
For this to be on the hypotenuse of the right triangle, we also have the following:
Note that the area of the equilateral triangle is given by , so we seek to minimize
. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:
Thus, the minimum we seek is simply , so the desired answer is
.
Solution 6
In the complex plane, let the vertices of the triangle be
and
Let
be one of the vertices, where
is real. A point on the line passing through
and
can be expressed in the form
We want the third vertex
to lie on the line through
and
which is the imaginary axis, so its real part is 0.
Since the small triangle is equilateral,
or
Then the real part of
is
Solving for
in terms of
we find
Then
so
so
This quadratic is minimized when
and the minimum is
so the smallest area of the equilateral triangle is
Solution 7
We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be and the point on the imaginary axis be
. Then, we see that
Now we switch back to Cartesian coordinates. The equation of the hypotenuse is
This means that the point
is on the line. Plugging the numbers in, we have
Now, we note that the side length of the equilateral triangle is
so it suffices to minimize that. By Cauchy-Schwarz, we have
Thus, the area of the smallest triangle is
so our desired answer is
.
(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 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 be the triangle with side lengths
and
.
We will think about this problem backwards, by constructing a triangle as large as possible (We will call it , for convenience) which is similar to
with vertices outside of a unit equilateral triangle
, such that each vertex of the equilateral triangle lies on a side of
. After we find the side lengths of
, we will use ratios to trace back towards the original problem.
First of all, let ,
, and
(These three angles are simply the angles of triangle
; out of these three angles,
is the smallest angle, and
is the largest angle). Then let us consider a point
inside
such that
,
, and
. Construct the circumcircles
and
of triangles
and
respectively.
From here, we will prove the lemma that if we choose points ,
, and
on circumcircles
and
respectively such that
,
, and
are collinear and
,
, and
are collinear, then
,
, and
must be collinear. First of all, if we let
, then
(by the properties of cyclic quadrilaterals),
(by adjacent angles),
(by cyclic quadrilaterals),
(adjacent angles), and
(cyclic quadrilaterals). Since
and
are supplementary,
,
, and
are collinear as desired. Hence,
has an inscribed equilateral triangle
.
In addition, now we know that all triangles (as described above) must be similar to triangle
, as
and
, so we have developed
similarity between the two triangles. Thus,
is the triangle similar to
which we were desiring. Our goal now is to maximize the length of
, in order to maximize the area of
, to achieve our original goal.
Note that, all triangles are similar to each other if
,
, and
are collinear. This is because
is constant, and
is also a constant value. Then we have
similarity between this set of triangles. To maximize
, we can instead maximize
, which is simply the diameter of
. From there, we can determine that
, and with similar logic,
,
, and
are perpendicular to
,
, and
respectively We have found our desired largest possible triangle
.
All we have to do now is to calculate , and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within
. First of all, we will prove that
. By the properties of cyclic quadrilaterals,
, which means that
. Now we will show that
. Note that, by cyclic quadrilaterals,
and
. Hence,
(since
), proving the aforementioned claim. Then, since
and
,
.
Now we calculate and
, which are simply the diameters of circumcircles
and
, respectively. By the extended law of sines,
and
.
We can now solve for with the law of cosines:
Now we will apply this discovery towards our original triangle . Since the ratio between
and the hypotenuse of
is
, the side length of the equilateral triangle inscribed within
must be
(as
is simply as scaled version of
, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within
is
, implying that the answer is
.
-Solution by TheBoomBox77
Solution 10
Let the right triangle's lower-left point be at . Notice the 2 other points will determine a unique equilateral triangle. Let 2 points be on the
-axis (
) and the
-axis (
) and label them
and
respectively. The third point (
) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of
and
.
1. Find the slope of and take the negative reciprocal of it to find the slope of the line containing
. Notice the line contains the midpoint of
so we can then have an equation of the line.
2. Let For
to be an equilateral triangle, the altitude from
to
must be
We then have two equations and two variables, so we can solve for 's coordinates.
We can find
Also, note that
must be on the hypotenuse of the triangle
We can plug in
and
as the coordinates of
, which simplifies to
We aim to minimize the side length of the triangle, which is Applying the Cauchy inequality gives us
From which we obtain Thus, the area of the triangle =
which leads to the answer
-hi_im_bob
Solution 11
The general solution to the minimal area is as following:
where and
are the two legs of the right triangle. In this particular case
and
. When we plug in these two values, we recover the correct answer of
.
The contour of the minimal area is plotted as a function of leg lengths
and
, as shown on the right hand side.
Note
The proof of the formula can be done in a similar fashion as Solution 4, where instead of using specific values, we define variables, and
in this case, as per the formula.
Solution 12 (Geometry)
Let
Then
midpoint
midpoint
circumcenter
Points
and
lies on bisectors of
and
so
The smallest area is
vladimir.shelomovskii@gmail.com, vvsss
Solution 13 (Kinematics+Geometry)
Let
Then
Let the required triangle with minimal sides be constructed.
Let us vary its position in an acceptable way, that is, we will perform a movement in which the vertices of the remain on the sides of the
Any motion of a solid plane figure can be considered as rotation around some point, the center of rotation. The speed of movement of any point is perpendicular to the segment from the center of rotation to this point. With an acceptable variation, the velocities of the vertices of the
are directed along the sides
Consequently, there is a point
located at the intersection of perpendiculars to the sides of
around which
rotates. The bases of the perpendiculars dropped from
to the sides of
form a regular
Therefore is the first isodynamic point of
It is known that
The points and
are concyclic,
is the diameter, so
Similarly,
The smallest area
is
vladimir.shelomovskii@gmail.com, vvsss
See Also
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 |
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