2015 AMC 12B Problems/Problem 19

Revision as of 19:02, 12 March 2015 by Jayjayliu (talk | contribs) (Solution 1)


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 $12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}$.

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS