Difference between revisions of "2015 AMC 10B Problems/Problem 19"
(→Solution 2) |
(→Solution 2) |
||
Line 55: | Line 55: | ||
==Solution 2== | ==Solution 2== | ||
− | 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 + \angle YZA + 90 - \angle WXB = 180</math>. Simplifying carefully, we get <math>\angle YZA = \angle WXB</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>. |
− | |||
− | < | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | </ | ||
==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
.
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.