Difference between revisions of "2015 AMC 12B Problems/Problem 19"

Revision as of 18:02, 1 February 2016

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 $\overline{XY}$ is parallel to $\overline{BX}$ 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 $\boxed{\textbf{(C)}\; 12+12\sqrt{2}}$ .

@@ Line 28: / Line 28: @@
 Since <math>a</math> cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by <math>(144-a^2)</math>, and arrive at <math>a = 6\sqrt{2}</math>. The length of other leg of the triangle must be <math>\sqrt{144-72} = 6\sqrt{2}</math>. Thus, the perimeter of the triangle is <math>12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.
+==Solution 3==
+In order to solve this problem, we can search for similar triangles. Begin by drawing triangle <math>ABC</math> and squares <math>ABXY</math> and <math>ACWZ</math>. Draw segments <math>\overline{YZ}</math> and <math>\overline{WX}</math>. Because we are given points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle, we can conclude that <math>WXYZ</math> forms a cyclic quadrilateral. Take <math>\overline{AC}</math> and extend it through a point <math>P</math> on <math>\overline{YZ}</math>. Now, we must do some angle chasing to prove that <math>\triangle WBX</math> is similar to <math>\triangle YAZ</math>.
+Let <math>\alpha</math> denote the measure of <math>\angle ABC</math>. Following this, <math>\angle BAC</math> measures <math>90 - \alpha</math>. By our construction, <math>\overline{CAP}</math> is a straight line, and we know <math>\angle YAB</math> is a right angle. Therefore, <math>\angle PAY</math> measures <math>\alpha</math>. Also, <math>\angle CAZ</math> is a right angle and thus, <math>\angle ZAP</math> is a right angle. Sum <math>\angle ZAP</math> and <math>\angle PAY</math> to find <math>\angle ZAY</math>, which measures <math>90 + \alpha</math>. We also know that <math>\angle WBY</math> measures <math>90 + \alpha</math>. Therefore, <math>\angle ZAY = \angle WBX</math>.
+Let <math>\beta</math> denote the measure of <math>\angle AZY</math>. It follows that <math>\angle WZY</math> measures <math>90 + \beta^\circ</math>. Because <math>WXYZ</math> is a cyclic quadrilateral, <math>\angle WZY + \angle YXW = 180^\circ</math>. Therefore, <math>\angle YXW</math> must measure <math>90 - \beta</math>, and <math>\angle BXW</math> must measure <math>\beta</math>. Therefore, <math>\angle AZY = \angle BXW</math>.
+<math>\angle ZAY = \angle WBX</math> and <math>\angle AZY = \angle BXW</math>, so <math>\triangle AZY \sim \triangle BXW</math>! Let <math>x = AC = WC</math>. By Pythagorean theorem, <math>BC = \sqrt{144-x^2}</math>. Now we have <math>WB = WC + BC = x + \sqrt{144-x^2}</math>, <math>BX = 12</math>, <math>YA = 12</math>, and <math>AZ = x</math>.
+We can set up an equation:
+<cmath>\frac{YA}{AZ} = \frac{WB}{BX}</cmath>
+<cmath>\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}</cmath>
+<cmath>144 = x^2 + x\sqrt{144-x^2}</cmath>
+<cmath>12^2 - x^2 = x\sqrt{144-x^2}</cmath>
+<cmath>12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4</cmath>
+<cmath>2x^4 - 3(12^2)x^2 + 12^4 = 0</cmath>
+<cmath>(2x^2 - 144)(x^2 - 144) = 0</cmath>
+Solving for <math>x</math>, we find that <math>x = 6\sqrt{2}</math> or <math>x = 12</math>, which we omit. The perimeter of the triangle is <math>12 + x + \sqrt{144-x^2}</math>. Plugging in <math>x = 6\sqrt{2}</math>, we get <math>\boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.
 ==See Also==
 {{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}
 {{MAA Notice}}

2015 AMC 12B (Problems • Answer Key • Resources)
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