Difference between revisions of "2015 AMC 10B Problems/Problem 19"
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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, you can find that <math>m\angle{ZYA}</math> = <math>m\angle{XWB}</math>. So, we know that <math>\triangle ZYA</math> is similar to <math>\triangle WBX</math>. Let <math>AC = b</math> and <math>AB = c</math>. So <math>BC = \sqrt{c^2-b^2}</math>. Now, we can use the proportion <math>\dfrac{c}{b} = \dfrac{b + \sqrt{c^2-b^2}}{c}</math>. Simplifying, we get <math>c^2 - b^2 = b \sqrt{c^2-b^2}</math>, which gives us <math>c^2 = 2b^2</math>, or <math>c = \sqrt{2}b</math>. Therefore, <math>\triangle ABC</math> is an isosceles right triangle, so <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, and the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</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, you can find that <math>m\angle{ZYA}</math> = <math>m\angle{XWB}</math>. So, we know that <math>\triangle ZYA</math> is similar to <math>\triangle WBX</math>. Let <math>AC = b</math> and <math>AB = c</math>. So <math>BC = \sqrt{c^2-b^2}</math>. Now, we can use the proportion <math>\dfrac{c}{b} = \dfrac{b + \sqrt{c^2-b^2}}{c}</math>. Simplifying, we get <math>c^2 - b^2 = b \sqrt{c^2-b^2}</math>, which gives us <math>c^2 = 2b^2</math>, or <math>c = \sqrt{2}b</math>. Therefore, <math>\triangle ABC</math> is an isosceles right triangle, so <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, and the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>. | ||
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==See Also== | ==See Also== |
Revision as of 17:27, 4 May 2019
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
Draw line segments and . Now we have cyclic quadrilateral . This means that opposite angles sum to . Therefore, . Simplifying carefully, we get . Similarly, you can find that = . So, we know that is similar to . Let and . So . Now, we can use the proportion . Simplifying, we get , which gives us , or . Therefore, is an isosceles right triangle, so , and the perimeter is .
~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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