Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
Line 55: Line 55:
 
==Solution 2==
 
==Solution 2==
  
−
Draw line segments <math>YZ</math> and <math>WX</math>. Now we have cyclic quadrilateral <math>WXYZ</math>. This means that opposite angles sum to 180 degrees. Therefore, <math>90 + angle{YZA} + 90 - angle{WXB = 180</math>. Simplifying carefully, we get <math>\angle{YZA} = </math>\angle{WXB}<math>. You can use the same method to find that </math>\angle{ZYA}<math> = </math>\angle{XWB}$. So, we know that  
+
Draw line segments <math>YZ</math> and <math>WX</math>. Now we have cyclic quadrilateral <math>WXYZ</math>. This means that opposite angles sum to 180 degrees. Therefore, <math>90 + \angle YZA + 90 - \angle WXB = 180</math>. Simplifying carefully, we get <math>\angle{YZA} = </math>\angle{WXB}<math>. You can use the same method to find that </math>\angle{ZYA}<math> = </math>\angle{XWB}$. So, we know that  
  
 
<asy>  
 
<asy>  

Revision as of 16:09, 4 May 2019

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 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

Draw line segments $YZ$ and $WX$. Now we have cyclic quadrilateral $WXYZ$. This means that opposite angles sum to 180 degrees. Therefore, $90 + \angle YZA + 90 - \angle WXB = 180$. Simplifying carefully, we get $\angle{YZA} =$\angle{WXB}$. You can use the same method to find that$\angle{ZYA}$=$\angle{XWB}$. So, we know that

[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]

See Also

2015 AMC 10B (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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_19&oldid=105656"