Difference between revisions of "2015 AMC 10B Problems/Problem 19"

Revision as of 09:25, 28 December 2021

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

The center of the circle lies on the intersection between the perpendicular bisectors of 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 $\boxed{\textbf{(C)}\ 12+12\sqrt{2}}$

~BakedPotato66

~LegionOfAvatars

Solution 3

Both solution 1 and 2 uses Pythagorean Theorem to prove $\triangle ABC$ is isosceles right triangle. I'm going to prove $\triangle ABC$ is isosceles right triangle without using Pythagorean Theorem. I will use geometry rotation.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */ draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle); draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); /* draw figures */ draw((3.46,0.96)--(3.44,-3.36)); draw((3.44,-3.36)--(8.02,-3.44)); draw((8.02,-3.44)--(3.46,0.96)); draw((3.46,0.96)--(-0.86,0.98)); draw((-0.86,0.98)--(-0.88,-3.34)); draw((-0.88,-3.34)--(3.44,-3.36)); draw((3.46,0.96)--(8.02,-3.44)); draw((8.02,-3.44)--(12.42,1.12)); draw((12.42,1.12)--(7.86,5.52)); draw((7.86,5.52)--(3.46,0.96)); draw((5.74,-1.24)--(-0.86,0.98)); draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4")); draw((5.74,-1.24)--(7.86,5.52)); draw((5.74,-1.24)--(10.14,3.32), linetype("4 4")); draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2")); draw((8.02,-3.44)--(-0.86,0.98)); draw((3.44,-3.36)--(7.86,5.52)); draw((3.44,-3.36)--(5.74,-1.24)); /* dots and labels */ dot((3.46,0.96),dotstyle); label("$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$", (5.52,-1.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

Let $O$ be the the midpoint of $AB$ . The perpendicular bisector of line $WZ$ and $XY$ will meet at $O$ . Thus $O$ is the center of the circle points $X$ , $Y$ , $Z$ , and $W$ lies on.

$\angle ZAC=\angle OAY=90^{\circ}$ , $\angle ZAC+\angle BAC=\angle OAY+\angle BAC$ , $\angle ZAB=\angle CAY$ , and $AZ=AC$ , $AB=AY$ , $\triangle AZB \cong \triangle ACY$ by $SAS$ , $BZ=YC$ . Because $AZ \perp AC$ , $\triangle ACY$ is a $90^{\circ}$ rotation about point $A$ of $\triangle AZB$ . So, $BZ \perp YC$ .

Because $OZ$ and $OY$ is the radius of $\odot O$ , $OZ=OY$ . Because $O$ is the midpoint of hypotenuse $AB$ , $OB=OC$ , $BZ=CY$ , $\triangle OBZ \cong \triangle OCY$ by $SSS$ . Because $BZ \perp CY$ , $\triangle OCY$ is a $90^{\circ}$ rotation about point $O$ of $\triangle OBZ$ . So, $OB \perp OC$ .

$\angle COB = 90^{\circ}$ , $OC=OB$ , $\triangle OCB$ is isosceles right triangle, $\angle ABC=\angle OBC=45^{\circ}$ . So, $\triangle ABC$ is isosceles right triangle.

Therefore, $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$ , the perimeter is $\boxed{\textbf{(C)}\ 12+12\sqrt{2}}$ .

~isabelchen

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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