Difference between revisions of "2015 AMC 10B Problems/Problem 19"
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− | 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>. Aha! 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>. | + | 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> | ||
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Revision as of 19:58, 6 January 2019
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
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
. [i]Aha[/i]! This means
is just an isosceles right triangle, so
, and thus the perimeter is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.