Difference between revisions of "2015 AMC 10B Problems/Problem 19"
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<math>\textbf{(A) }12+9\sqrt{3}\qquad\textbf{(B) }18+6\sqrt{3}\qquad\textbf{(C) }12+12\sqrt{2}\qquad\textbf{(D) }30\qquad\textbf{(E) }32</math> | <math>\textbf{(A) }12+9\sqrt{3}\qquad\textbf{(B) }18+6\sqrt{3}\qquad\textbf{(C) }12+12\sqrt{2}\qquad\textbf{(D) }30\qquad\textbf{(E) }32</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | The center of the circle lies on the perpendicular bisectors of both chords <math>ZW</math> and <math>YX</math>. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be <math>O</math>. Draw perpendiculars to <math>ZW</math> and <math>YX</math> from <math>O</math>, and connect <math>OZ</math> and <math>OY</math>. <math>OY^2=6^2+12^2=180</math>. Let <math>AC=a</math> and <math>BC=b</math>. Then <math>\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180</math>. Simplifying this gives <math>\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180</math>. But by | + | The center of the circle lies on the perpendicular bisectors of both chords <math>ZW</math> and <math>YX</math>. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be <math>O</math>. Draw perpendiculars to <math>ZW</math> and <math>YX</math> from <math>O</math>, and connect <math>OZ</math> and <math>OY</math>. <math>OY^2=6^2+12^2=180</math>. Let <math>AC=a</math> and <math>BC=b</math>. Then <math>\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180</math>. Simplifying this gives <math>\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180</math>. But by Pythagorean Theorem on <math>\triangle ABC</math>, we know <math>a^2+b^2=144</math>, because <math>AB=12</math>. Thus <math>\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36</math>. So our equation simplifies further to <math>a^2+ab=144</math>. However <math>a^2+b^2=144</math>, so <math>a^2+ab=a^2+b^2</math>, which means <math>ab=b^2</math>, or <math>a=b</math>. <i>Aha</i>! This means <math>\triangle ABC</math> is just an isosceles right triangle, so <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, and thus the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>. |
<asy> | <asy> | ||
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
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label("$O$", (5.52,-1.82), NE * labelscalefactor); | label("$O$", (5.52,-1.82), NE * labelscalefactor); | ||
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
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</asy> | </asy> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>AC = b</math> and <math>BC = a</math> (and we're given that <math>AB=12</math>). Draw line segments <math>YZ</math> and <math>WX</math>. Now we have cyclic quadrilateral <math>WXYZ.</math> | ||
+ | |||
+ | This means that opposite angles sum to <math>180^{\circ}</math>. Therefore, <math>90 + m\angle YZA + 90 - m\angle WXB = 180</math>. Simplifying carefully, we get <math>m\angle YZA = m\angle WXB</math>. Similarly, <math>m\angle{ZYA}</math> = <math>m\angle{XWB}</math>. | ||
+ | |||
+ | That means <math>\triangle ZYA \sim \triangle XWB</math>. | ||
+ | |||
+ | Setting up proportions, | ||
+ | <math>\dfrac{b}{12}=\dfrac{12}{a+b}.</math> | ||
+ | Cross-multiplying we get: | ||
+ | <math>b^2+ab=12^2</math> | ||
+ | |||
+ | But also, by Pythagoras, | ||
+ | <math>b^2+a^2=12^2</math>, so <math>ab=a^2 \Rightarrow a=b</math> | ||
+ | |||
+ | Therefore, <math>\triangle ABC</math> is an isosceles right triangle. <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, so the perimeter is <cmath>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</cmath> | ||
+ | |||
+ | ~BakedPotato66 | ||
+ | |||
+ | ~LegionOfAvatars | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2015|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Geometry Problems]] |
Revision as of 15:58, 12 October 2020
Contents
Problem
In , and . Squares and are constructed outside of the triangle. The points , and lie on a circle. What is the perimeter of the triangle?
Solution 1
The center of the circle lies on the perpendicular bisectors of both chords and . Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be . Draw perpendiculars to and from , and connect and . . Let and . Then . Simplifying this gives . But by Pythagorean Theorem on , we know , because . Thus . So our equation simplifies further to . However , so , which means , or . Aha! This means is just an isosceles right triangle, so , and thus the perimeter is .
Solution 2
Let and (and we're given that ). Draw line segments and . Now we have cyclic quadrilateral
This means that opposite angles sum to . Therefore, . Simplifying carefully, we get . Similarly, = .
That means .
Setting up proportions, Cross-multiplying we get:
But also, by Pythagoras, , so
Therefore, is an isosceles right triangle. , so the perimeter is
~BakedPotato66
~LegionOfAvatars
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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