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

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Now, we can see that <math>\angle BAC=45^\circ</math> to complete the full <math>360^\circ</math>. Therefore, <math>\triangle ABC</math> is and isosceles right triangle, and <math>AC=BC=6\sqrt2</math>. So Our answer is <math>6\sqrt2 + 6\sqrt 2 + 12 = \boxed{\textbf{(C) }12+12\sqrt{2}}</math>.
 
Now, we can see that <math>\angle BAC=45^\circ</math> to complete the full <math>360^\circ</math>. Therefore, <math>\triangle ABC</math> is and isosceles right triangle, and <math>AC=BC=6\sqrt2</math>. So Our answer is <math>6\sqrt2 + 6\sqrt 2 + 12 = \boxed{\textbf{(C) }12+12\sqrt{2}}</math>.
 
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==Solution 5==
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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 <math>WXYZ</math> is cyclic. Then we see that <math>\angle W = \angle X</math> making extensions of <math>\overline{WA}</math> and <math>\overline{XZ}</math> diagonals of <math>WXYZ</math>. Let <math>x = \overline{AC}=\overline{AW}</math>. We can see that <math>\overline{AY} = 12\sqrt{2}</math> and <math>\overline{AZ} = x\sqrt{2}</math>. Thus <math>\overline{YZ}^2 = x^2 + (x+12\sqrt{2})^2 = 2x^2 + 24x\sqrt{2} + 288</math>. Let <math>\angle ZAY = \alpha</math>. Then we can see that <math>\angle ZAC</math> and <math>\angle BAY</math> are <math>45</math> degrees, making <math>\angle ZAY</math> <math>\alpha + 90</math> degrees.
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We can verify with the cosine addition identity that <math>-\cos{\alpha + 90 degree} = \sin{\alpha}</math> (knowing that <math>\sin{\theta + 90 degree} = \cos{\theta}</math> motivates this). By law of cosine's, <math>\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}</math>. Since <math>\angle{A}</math> is opposite <math>\overline{BC}</math>, <math>\sin{\alpha} = \frac{\sqrt{144-x^2}}{12}</math>. Thus <math>\overline{YZ}^2 = 288 + 2x^2 + 4x\sqrt{144-x^2}</math>.
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Setting the first paragraph's <math>\overline{YZ}^2</math> equal to the second:
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<cmath>288 + x^2 + 4x\sqrt{144-x^2}2x^2 + 24x\sqrt{2} + 288</cmath>
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==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 20:05, 7 November 2021

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 5

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: \[288 + x^2 + 4x\sqrt{144-x^2}2x^2 + 24x\sqrt{2} + 288\]


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

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