Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 12B Problems/Problem 19"
(→‎Solution 4)
Line 52: Line 52:
 
We claim that <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles right triangle.
 
We claim that <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles right triangle.
  
−
Proof: If <math>\triangle ACB</math> is a right triangle, then <math>\angle WAY=180º</math>. Therefore, <math>W</math>, <math>A</math>, and <math>Y</math> are collinear. Since <math>WY</math> and <math>YX</math> form a right angle, <math>WX</math> is the diameter of the circumcircle of <math>\triangle WYX</math>. Similarly, <math>Z</math>, <math>A</math>, and <math>X</math> are collinear, and <math>ZX</math> forms a right angle with <math>ZW</math>. Thus, <math>WX</math> is also the diameter of the circumcircle of <math>\triangle WZX</math>. Therefore, since <math>\triangle WYX</math> and <math>\triangle WZX</math> share a circumcircle, <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles triangle.
+
''Proof:'' If <math>\triangle ACB</math> is a right triangle, then <math>\angle WAY=180º</math>. Therefore, <math>W</math>, <math>A</math>, and <math>Y</math> are collinear. Since <math>WY</math> and <math>YX</math> form a right angle, <math>WX</math> is the diameter of the circumcircle of <math>\triangle WYX</math>. Similarly, <math>Z</math>, <math>A</math>, and <math>X</math> are collinear, and <math>ZX</math> forms a right angle with <math>ZW</math>. Thus, <math>WX</math> is also the diameter of the circumcircle of <math>\triangle WZX</math>. Therefore, since <math>\triangle WYX</math> and <math>\triangle WZX</math> share a circumcircle, <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles triangle.
  
−
If <math>\triangle ACB</math> is isosceles, then its legs have length <math>6\sqrt{2}</math>. The perimeter of <math>\triangle ACB</math> is <math>\textbf{(C) }12+12\sqrt{2}</math>.
+
If <math>\triangle ACB</math> is isosceles, then its legs have length <math>6\sqrt{2}</math>. The perimeter of <math>\triangle ACB</math> is <math>\boxed{\textbf{(C) }12+12\sqrt{2}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}
 
{{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:10, 16 December 2017

Problem

In $\triangle ABC$, $\angle C = 90^\circ$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X$, $Y$, $Z$, and $W$ lie on a circle. What is the perimeter of the triangle?

$\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32$

Solution 1

First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of $WZ$ and $XY$ and finding their intersection point. This point happens to be the midpoint of $AB$, the hypotenuse. Let this point be $M$. To find the radius, determine $MY$, where $MY^{2} = MA^2 + AY^2$, $MA = \frac{12}{2} = 6$, and $AY = AB = 12$. Thus, the radius $=r =MY = 6\sqrt5$.

Next we let $AC = b$ and $BC = a$. Consider the right triangle $ACB$ first. Using the pythagorean theorem, we find that $a^2 + b^2 = 12^2 = 144$. Next, we let $M'$ to be the midpoint of $WZ$, and we consider right triangle $ZM'M$. By the pythagorean theorem, we have that $\left(\frac{b}{2}\right)^2 + \left(b + \frac{a}{2}\right)^2 = r^2 = 180$. Expanding this equation, we get that

\[\frac{1}{4}(a^2+b^2) + b^2 + ab = 180\] \[\frac{144}{4} + b^2 + ab = 180\] \[b^2 + ab = 144 = a^2 + b^2\] \[ab = a^2\] \[b = a\]

This means that $ABC$ is a 45-45-90 triangle, so $a = b = \frac{12}{\sqrt2} = 6\sqrt2$. Thus the perimeter is $a + b + AB = 12\sqrt2 + 12$ which is answer $\boxed{\textbf{(C)}\; 12 + 12\sqrt2}$. image needed

Solution 2

The center of the circle on which $X$, $Y$, $Z$, and $W$ lie must be equidistant from each of these four points. Draw the perpendicular bisectors of $\overline{XY}$ and of $\overline{WZ}$. Note that the perpendicular bisector of $\overline{WZ}$ is parallel to $\overline{CW}$ and passes through the midpoint of $\overline{AC}$. Therefore, the triangle that is formed by $A$, the midpoint of $\overline{AC}$, and the point at which this perpendicular bisector intersects $\overline{AB}$ must be similar to $\triangle ABC$, and the ratio of a side of the smaller triangle to a side of $\triangle ABC$ is 1:2. Consequently, the perpendicular bisector of $\overline{XY}$ passes through the midpoint of $\overline{AB}$. The perpendicular bisector of $\overline{WZ}$ must include the midpoint of $\overline{AB}$ as well. Since all points on a perpendicular bisector of any two points $M$ and $N$ are equidistant from $M$ and $N$, the center of the circle must be the midpoint of $\overline{AB}$.

Now the distance between the midpoint of $\overline{AB}$ and $Z$, which is equal to the radius of this circle, is $\sqrt{12^2 + 6^2} = \sqrt{180}$. Let $a=AC$. Then the distance between the midpoint of $\overline{AB}$ and $Y$, also equal to the radius of the circle, is given by $\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}$ (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have

\[\left(\frac{a}{2}\right)^2+\left(a+\frac{\sqrt{144-a^2}}{2}\right)^{2} = 180\] \[144 - a^2 = a\sqrt{144-a^2}\] \[(144-a^2)^2 = a^2(144-a^2)\]

Since $a$ cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by $(144-a^2)$, and arrive at $a = 6\sqrt{2}$. The length of other leg of the triangle must be $\sqrt{144-72} = 6\sqrt{2}$. Thus, the perimeter of the triangle is $12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}$.

Solution 3

In order to solve this problem, we can search for similar triangles. Begin by drawing triangle $ABC$ and squares $ABXY$ and $ACWZ$. Draw segments $\overline{YZ}$ and $\overline{WX}$. Because we are given points $X$, $Y$, $Z$, and $W$ lie on a circle, we can conclude that $WXYZ$ forms a cyclic quadrilateral. Take $\overline{AC}$ and extend it through a point $P$ on $\overline{YZ}$. Now, we must do some angle chasing to prove that $\triangle WBX$ is similar to $\triangle YAZ$.

Let $\alpha$ denote the measure of $\angle ABC$. Following this, $\angle BAC$ measures $90 - \alpha$. By our construction, $\overline{CAP}$ is a straight line, and we know $\angle YAB$ is a right angle. Therefore, $\angle PAY$ measures $\alpha$. Also, $\angle CAZ$ is a right angle and thus, $\angle ZAP$ is a right angle. Sum $\angle ZAP$ and $\angle PAY$ to find $\angle ZAY$, which measures $90 + \alpha$. We also know that $\angle WBY$ measures $90 + \alpha$. Therefore, $\angle ZAY = \angle WBX$.

Let $\beta$ denote the measure of $\angle AZY$. It follows that $\angle WZY$ measures $90 + \beta^\circ$. Because $WXYZ$ is a cyclic quadrilateral, $\angle WZY + \angle YXW = 180^\circ$. Therefore, $\angle YXW$ must measure $90 - \beta$, and $\angle BXW$ must measure $\beta$. Therefore, $\angle AZY = \angle BXW$.

$\angle ZAY = \angle WBX$ and $\angle AZY = \angle BXW$, so $\triangle AZY \sim \triangle BXW$! Let $x = AC = WC$. By Pythagorean theorem, $BC = \sqrt{144-x^2}$. Now we have $WB = WC + BC = x + \sqrt{144-x^2}$, $BX = 12$, $YA = 12$, and $AZ = x$. We can set up an equation:

\[\frac{YA}{AZ} = \frac{WB}{BX}\] \[\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}\] \[144 = x^2 + x\sqrt{144-x^2}\] \[12^2 - x^2 = x\sqrt{144-x^2}\] \[12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4\] \[2x^4 - 3(12^2)x^2 + 12^4 = 0\] \[(2x^2 - 144)(x^2 - 144) = 0\]

Solving for $x$, we find that $x = 6\sqrt{2}$ or $x = 12$, which we omit. The perimeter of the triangle is $12 + x + \sqrt{144-x^2}$. Plugging in $x = 6\sqrt{2}$, we get $\boxed{\textbf{(C)}\; 12+12\sqrt{2}}$.

Solution 4

We claim that $X$, $Y$, $Z$, and $W$ lie on a circle if $\triangle ACB$ is an isosceles right triangle.

Proof: If $\triangle ACB$ is a right triangle, then $\angle WAY=180º$. Therefore, $W$, $A$, and $Y$ are collinear. Since $WY$ and $YX$ form a right angle, $WX$ is the diameter of the circumcircle of $\triangle WYX$. Similarly, $Z$, $A$, and $X$ are collinear, and $ZX$ forms a right angle with $ZW$. Thus, $WX$ is also the diameter of the circumcircle of $\triangle WZX$. Therefore, since $\triangle WYX$ and $\triangle WZX$ share a circumcircle, $X$, $Y$, $Z$, and $W$ lie on a circle if $\triangle ACB$ is an isosceles triangle.

If $\triangle ACB$ is isosceles, then its legs have length $6\sqrt{2}$. The perimeter of $\triangle ACB$ is $\boxed{\textbf{(C) }12+12\sqrt{2}}$.

See Also

2015 AMC 12B (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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_19&oldid=88939"