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

Latest revision as of 21:42, 16 June 2023

Problem

In $\triangle ABC$ , $\angle C = 90^\circ$ and $AB = 12$ . Squares $ABXY$ and $CBWZ$ 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

$[asy] pair A,B,C,M,E,W,Z,X,Y; A=(2,0); B=(0,2); C=(0,0); M=(A+B)/2; W=(-2,2); Z=(-2,-0); X=(2,4); Y=(4,2); E=(W+Z)/2; draw(A--B--C--cycle); draw(W--B--C--Z--cycle); draw(A--B--X--Y--cycle); dot(M); dot(E); label("W",W,NW); label("Z",Z,SW); label("C",C,S); label("A",A,S); label("B",B,N); label("X",X,NE); label("Y",Y,SE); label("E",E,1.5*plain.W); label("M",M,NE); draw(circle(M,sqrt(10))); [/asy]$ 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 $a^2 + b^2 = 12^2 = 144$ . $[asy] pair A,B,C,M,E,W,Z,X,Y; A=(2,0); B=(0,2); C=(0,0); M=(A+B)/2; W=(-2,2); Z=(-2,-0); X=(2,4); Y=(4,2); E=(W+Z)/2; draw(A--B--C--cycle); draw(W--B--C--Z--cycle); draw(A--B--X--Y--cycle); dot(M); dot(E); label("W",W,NW); label("Z",Z,SW); label("C",C,S); label("A",A,S); label("B",B,N); label("X",X,NE); label("Y",Y,SE); label("E",E,1.5*plain.W); label("M",M,NE); draw(circle(M,sqrt(10))); draw(E--Z--M--cycle,dashed); [/asy]$ Now, we let $E$ be the midpoint of $WZ$ , and we consider right triangle $ZEM$ . By the Pythagorean theorem, we have that $\left(\frac{a}{2}\right)^2 + \left(a + \frac{b}{2}\right)^2 = r^2 = 180$ . Expanding this equation, we get that

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

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 $\boxed{\textbf{(C)}\; 12 + 12\sqrt2}$ .

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 $\boxed{\textbf{(C)}\; 12+12\sqrt{2}}$ .

Alternatively, let $BC = S_1, AB = S_2$ and $AC = x$ . Because $\frac{WB}{BX} = \frac{AY}{AZ}$ , we get that $S_2^2 = S_1^2 + S_1x$ . $S_1 = x$ satisfies the equation because of Pythagorean theorem, so $\triangle ABC$ is right isosceles.

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 an isosceles 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 $\boxed{\textbf{(C) }12+12\sqrt{2}}$ .

Solution 5

Note that because $\angle WZA=\angle AYX=90^\circ$ , these two angles inscribe the semicircle defined by diameter $WX$ . Since $A$ lies on line $ZA$ , and $\angle XAY=45^\circ$ (because $ABXY$ is a square), we can find that $\angle ZAY= 180^\circ - 45^\circ = 135^\circ$ .

Now, we can see that $\angle BAC=45^\circ$ to complete the full $360^\circ$ . Therefore, $\triangle ABC$ is and isosceles right triangle, and $AC=BC=6\sqrt2$ . So Our answer is $6\sqrt2 + 6\sqrt 2 + 12 = \boxed{\textbf{(C) }12+12\sqrt{2}}$ . .

Solution 6

We see a circle and little information of position of the shapes inside the triangle, so we think of things associated with circles and think of cyclic quads. We then notice that quadrilateral $WXYZ$ is cyclic. Then we see that $\angle W = \angle X$ making extensions of $\overline{WA}$ and $\overline{XZ}$ diagonals of $WXYZ$ . Let $x = \overline{AC}=\overline{AW}$ . We can see that $\overline{AY} = 12\sqrt{2}$ and $\overline{AZ} = x\sqrt{2}$ . Thus $\overline{YZ}^2 = x^2 + (x+12\sqrt{2})^2 = 2x^2 + 24x\sqrt{2} + 288$ . Let $\angle ZAY = \alpha$ . Then we can see that $\angle ZAC$ and $\angle BAY$ are $45$ degrees, making $\angle ZAY$ $\alpha + 90$ degrees.

We can verify with the cosine addition identity that $-\cos{\alpha + 90 degree} = \sin{\alpha}$ (knowing that $\sin{\theta + 90 degree} = \cos{\theta}$ motivates this). By law of cosine's, $\overline{YZ}^2 = (12\sqrt{2})^2 + x\sqrt{2})^2 - 2\cdot 12\sqrt{2} \cdot x\sqrt{2}cos{\alpha + 90 degree} = 288 + x^2 + 48cos{\alpha + 90 degree} = 288 + 2x^2 + 48x\sin{\alpha}$ . Since $\angle{A}$ is opposite $\overline{BC}$ , $\sin{\alpha} = \frac{\sqrt{144-x^2}}{12}$ . Thus $\overline{YZ}^2 = 288 + 2x^2 + 4x\sqrt{144-x^2}$ .

Setting the first paragraph's $\overline{YZ}^2$ equal to the second: $288 + x^2 + 4x\sqrt{144-x^2}2x^2 + 24x\sqrt{2} + 288$

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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