https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Asv2010&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T13:06:33ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_12&diff=518512013 AIME I Problems/Problem 122013-03-18T15:48:15Z<p>Asv2010: /* Solution */</p>
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<div>== Problem 12 ==<br />
Let <math>\bigtriangleup PQR</math> be a triangle with <math>\angle P = 75^\circ</math> and <math>\angle Q = 60^\circ</math>. A regular hexagon <math>ABCDEF</math> with side length 1 is drawn inside <math>\triangle PQR</math> so that side <math>\overline{AB}</math> lies on <math>\overline{PQ}</math>, side <math>\overline{CD}</math> lies on <math>\overline{QR}</math>, and one of the remaining vertices lies on <math>\overline{RP}</math>. There are positive integers <math>a, b, c, </math> and <math>d</math> such that the area of <math>\triangle PQR</math> can be expressed in the form <math>\frac{a+b\sqrt{c}}{d}</math>, where <math>a</math> and <math>d</math> are relatively prime, and c is not divisible by the square of any prime. Find <math>a+b+c+d</math>.<br />
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== Solution ==<br />
First, find that <math>\angle R = 45^\circ</math>.<br />
Draw <math>ABCDEF</math>. Now draw <math>\bigtriangleup PQR</math> around <math>ABCDEF</math> such that <math>Q</math> is adjacent to <math>C</math> and <math>D</math>. The height of <math>ABCDEF</math> is <math>\sqrt{3}</math>, so the length of base <math>QR</math> is <math>2+\sqrt{3}</math>. Let the equation of <math>\overline{RP}</math> be <math>y = x</math>. Then, the equation of <math>\overline{PQ}</math> is <math>y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3</math>. Solving the two equations gives <math>y = x = \frac{\sqrt{3} + 3}{2}</math>. The area of <math>\bigtriangleup PQR</math> is <math>\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}</math>. <math>a + b + c + d = 9 + 5 + 3 + 4 = 21</math><br />
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== See also ==<br />
{{AIME box|year=2013|n=I|num-b=11|num-a=13}}</div>Asv2010https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_18&diff=479262008 AMC 10A Problems/Problem 182012-08-17T16:10:38Z<p>Asv2010: /* Solution 4 */</p>
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<div>==Problem==<br />
A [[right triangle]] has [[perimeter]] <math>32</math> and area <math>20</math>. What is the length of its [[hypotenuse]]?<br />
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<math>\mathrm{(A)}\ \frac{57}{4}\qquad\mathrm{(B)}\ \frac{59}{4}\qquad\mathrm{(C)}\ \frac{61}{4}\qquad\mathrm{(D)}\ \frac{63}{4}\qquad\mathrm{(E)}\ \frac{65}{4}</math><br />
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__TOC__<br />
==Solution==<br />
=== Solution 1 ===<br />
Let the legs of the triangle have lengths <math>a,b</math>. Then, by the [[Pythagorean Theorem]], the length of the hypotenuse is <math>\sqrt{a^2+b^2}</math>, and the area of the triangle is <math>\frac 12 ab</math>. So we have the two equations<br />
<center><math>\begin{align}<br />
a+b+\sqrt{a^2+b^2} &= 32 \\<br />
\frac{1}{2}ab &= 20<br />
\end{align}</math></center><br />
Re-arranging the first equation and squaring, <br />
<center><math>\begin{align*}<br />
\sqrt{a^2+b^2} &= 32-(a+b)\\<br />
a^2 + b^2 &= 32^2 - 64(a+b) + (a+b)^2\\<br />
a^2 + b^2 + 64(a+b) &= a^2 + b^2 + 2ab + 32^2\\<br />
a+b &= \frac{2ab+32^2}{64}\end{align*}</math></center><br />
From <math>(2)</math> we have <math>2ab = 80</math>, so<br />
<center><math>a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.</math></center><br />
The length of the hypotenuse is <math>p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\ \mathrm{(B)}</math>.<br />
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=== Solution 2 ===<br />
From the formula <math>A = rs</math>, where <math>A</math> is the area of a triangle, <math>r</math> is its [[inradius]], and <math>s</math> is the [[semiperimeter]], we can find that <math>r = \frac{20}{32/2} = \frac{5}{4}</math>. It is known that in a right triangle, <math>r = s - h</math>, where <math>h</math> is the hypotenuse, so <math>h = 16 - \frac{5}{4} = \frac{59}{4}</math>.<br />
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=== Solution 3 ===<br />
From the problem, we know that <br />
<center><math>\begin{align*}<br />
a+b+c &= 32 \\<br />
2ab &= 80. \\<br />
\end{align*}</math></center><br />
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Subtracting <math>c</math> from both sides of the first equation and squaring both sides, we get<br />
<center><math>\begin{align*}<br />
(a+b)^2 &= (32 - c)^2\\<br />
a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\<br />
\end{align*}</math></center><br />
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Now we substitute in <math>a^2 + b^2 = c^2</math> as well as <math>2ab = 80</math> into the equation to get<br />
<center><math>\begin{align*}<br />
80 &= 1024 - 64c\\<br />
c &= \frac{944}{64}.<br />
\end{align*}</math></center><br />
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Further simplification yields the result of <math>\frac{59}{4}</math>.<br />
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=== Solution 4 ===<br />
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Let <math>a</math> and <math>b</math> be the legs of the triangle, and <math>c</math> the hypotenuse.<br />
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Since the area is 20, we have <math>\frac{1}{2}ab = 20 => ab=40</math>.<br />
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Since the perimeter is 32, we have <math>a + b + c = 32</math>.<br />
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The Pythagorean Theorem gives <math>c^2 = a^2 + b^2</math>. <br />
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This gives us three equations with three variables:<br />
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<center><math>ab = 40 \\<br />
a + b + c = 32 \\<br />
c^2 = a^2 + b^2</math></center><br />
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Rewrite equation 3 as <math>c^2 = (a+b)^2 - 2ab</math>.<br />
Substitute in equations 1 and 2 to get <math>c^2 = (32-c)^2 - 80</math>.<br />
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<center><math>c^2 = (32-c)^2 - 80 \\<br />
c^2 = 1024 - 64c + c^2 - 80 \\<br />
64c = 944 \\<br />
c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4} </math>. </center><br />
The answer is choice (B).<br />
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==See also==<br />
{{AMC10 box|year=2008|ab=A|num-b=17|num-a=19}}<br />
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[[Category:Introductory Geometry Problems]]</div>Asv2010