2015 AMC 10B Problems/Problem 19


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

The center of the circle lies on the perpendicular bisectors of both chords $ZW$ and $YX$. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be $O$. Draw perpendiculars to $ZW$ and $YX$ from $O$, and connect $OZ$ and $OY$. $OY^2=6^2+12^2=180$. Let $AC=a$ and $BC=b$. Then $\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180$. Simplifying this gives $\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180$. But by Pythagorean Theorem on $\triangle ABC$, we know $a^2+b^2=144$, because $AB=12$. Thus $\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36$. So our equation simplifies further to $a^2+ab=144$. However $a^2+b^2=144$, so $a^2+ab=a^2+b^2$, which means $ab=b^2$, or $a=b$. Aha! This means $\triangle ABC$ is just an isosceles right triangle, so $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$, and thus the perimeter is $\boxed{\textbf{(C)}\ 12+12\sqrt{2}}$. [asy]   /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.5cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3;  /* image dimensions */   draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle);  draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle);   /* draw figures */ draw((3.46,0.96)--(3.44,-3.36));  draw((3.44,-3.36)--(8.02,-3.44));  draw((8.02,-3.44)--(3.46,0.96));  draw((3.46,0.96)--(-0.86,0.98));  draw((-0.86,0.98)--(-0.88,-3.34));  draw((-0.88,-3.34)--(3.44,-3.36));  draw((3.46,0.96)--(8.02,-3.44));  draw((8.02,-3.44)--(12.42,1.12));  draw((12.42,1.12)--(7.86,5.52));  draw((7.86,5.52)--(3.46,0.96));  draw((5.74,-1.24)--(-0.86,0.98));  draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4"));  draw((5.74,-1.24)--(7.86,5.52));  draw((5.74,-1.24)--(10.14,3.32), linetype("4 4"));  draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2"));   /* dots and labels */ dot((3.46,0.96),dotstyle);  label("$A$", (3.2,1.06), NE * labelscalefactor);  dot((3.44,-3.36),dotstyle);  label("$C$", (3.14,-3.86), NE * labelscalefactor);  dot((8.02,-3.44),dotstyle);  label("$B$", (8.06,-3.8), NE * labelscalefactor);  dot((-0.86,0.98),dotstyle);  label("$Z$", (-1.34,1.12), NE * labelscalefactor);  dot((-0.88,-3.34),dotstyle);  label("$W$", (-1.48,-3.54), NE * labelscalefactor);  dot((12.42,1.12),dotstyle);  label("$X$", (12.5,1.24), NE * labelscalefactor);  dot((7.86,5.52),dotstyle);  label("$Y$", (7.94,5.64), NE * labelscalefactor);  dot((5.74,-1.24),dotstyle);  label("$O$", (5.52,-1.82), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  [/asy]

Solution 2

Let $AC = b$ and $BC = a$ (and we're given that $AB=12$). Draw line segments $YZ$ and $WX$. Now we have cyclic quadrilateral $WXYZ.$

This means that opposite angles sum to $180^{\circ}$. Therefore, $90 + m\angle YZA + 90 - m\angle WXB = 180$. Simplifying carefully, we get $m\angle YZA = m\angle WXB$. Similarly, $m\angle{ZYA}$ = $m\angle{XWB}$.

That means $\triangle ZYA \sim \triangle XWB$.

Setting up proportions, $\dfrac{b}{12}=\dfrac{12}{a+b}.$ Cross-multiplying we get: $b^2+ab=12^2$

But also, by Pythagoras, $b^2+a^2=12^2$, so $ab=a^2 \Rightarrow a=b$

Therefore, $\triangle ABC$ is an isosceles right triangle. $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$, so the perimeter is \[\boxed{\textbf{(C)}\ 12+12\sqrt{2}}\]



See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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