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

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==Solution 1==
 
==Solution 1==
−
The center of the circle lies on the perpendicular bisectors of both chords <math>ZW</math> and <math>YX</math>. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be <math>O</math>. Draw perpendiculars to <math>ZW</math> and <math>YX</math> from <math>O</math>, and connect <math>OZ</math> and <math>OY</math>. <math>OY^2=6^2+12^2=180</math>. Let <math>AC=a</math> and <math>BC=b</math>. Then <math>\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180</math>. Simplifying this gives <math>\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180</math>. But by Pythagorean Theorem on <math>\triangle ABC</math>, we know <math>a^2+b^2=144</math>, because <math>AB=12</math>. Thus <math>\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36</math>. So our equation simplifies further to <math>a^2+ab=144</math>. However <math>a^2+b^2=144</math>, so <math>a^2+ab=a^2+b^2</math>, which means <math>ab=b^2</math>, or <math>a=b</math>. <i>Aha</i>! This means <math>\triangle ABC</math> is just an isosceles right triangle, so <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, and thus the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>.
+
The center of the circle lies on the intersection between the perpendicular bisectors of chords <math>ZW</math> and <math>YX</math>. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be <math>O</math>. Draw perpendiculars to <math>ZW</math> and <math>YX</math> from <math>O</math>, and connect <math>OZ</math> and <math>OY</math>. <math>OY^2=6^2+12^2=180</math>. Let <math>AC=a</math> and <math>BC=b</math>. Then <math>\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180</math>. Simplifying this gives <math>\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180</math>. But by Pythagorean Theorem on <math>\triangle ABC</math>, we know <math>a^2+b^2=144</math>, because <math>AB=12</math>. Thus <math>\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36</math>. So our equation simplifies further to <math>a^2+ab=144</math>. However <math>a^2+b^2=144</math>, so <math>a^2+ab=a^2+b^2</math>, which means <math>ab=b^2</math>, or <math>a=b</math>. <i>Aha</i>! This means <math>\triangle ABC</math> is just an isosceles right triangle, so <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, and thus the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>.
 
<asy>  
 
<asy>  
 
  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
 
  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
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 <math>180^{\circ} degrees. Therefore, </math>90 + \angle YZA + 90 - \angle WXB = 180<math>. Simplifying carefully, we get </math>\angle YZA = \angle WXB<math>. You can use the same method to find that </math>\angle{ZYA}<math> = </math>\angle{XWB}$. So, we know that
+
Let <math>AC = b</math> and <math>BC = a</math> (and we're given that <math>AB=12</math>). 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 <math>180^{\circ}</math>. Therefore, <math>90 + m\angle YZA + 90 - m\angle WXB = 180</math>. Simplifying carefully, we get <math>m\angle YZA = m\angle WXB</math>. Similarly, <math>m\angle{ZYA}</math> = <math>m\angle{XWB}</math>.
 +
 
 +
That means <math>\triangle ZYA \sim \triangle XWB</math>.  
 +
 
 +
Setting up proportions,  
 +
<math>\dfrac{b}{12}=\dfrac{12}{a+b}.</math>
 +
Cross-multiplying we get:
 +
<math>b^2+ab=12^2</math>
 +
 
 +
But also, by Pythagoras,
 +
<math>b^2+a^2=12^2</math>, so <math>ab=a^2 \Rightarrow a=b</math>
 +
 
 +
Therefore, <math>\triangle ABC</math> is an isosceles right triangle. <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, so the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>
 +
 
 +
~BakedPotato66
 +
 
 +
~LegionOfAvatars
 +
 
 +
==Solution 3==
 +
 
 +
Both solution 1 and 2 uses Pythagorean Theorem to prove <math>\triangle ABC</math> is isosceles right triangle. I'm going to prove <math>\triangle ABC</math> is isosceles right triangle without using Pythagorean Theorem. I will use geometry rotation.
  
 
<asy>  
 
<asy>  
Line 64: Line 86:
 
pen dotstyle = black; /* point style */  
 
pen dotstyle = black; /* point style */  
 
real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3;  /* image dimensions */
 
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)--(3.44,-3.36)--(8.02,-3.44)--cycle);  
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draw((5.74,-1.24)--(10.14,3.32), linetype("4 4"));  
 
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(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 */
 
  /* dots and labels */
 
dot((3.46,0.96),dotstyle);  
 
dot((3.46,0.96),dotstyle);  
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  
 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  
 
</asy>
 
</asy>
 +
 +
Let <math>O</math> be the the midpoint of <math>AB</math>. The perpendicular bisector of line <math>WZ</math> and <math>XY</math> will meet at <math>O</math>. Thus <math>O</math> is the center of the circle points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on.
 +
 +
<math>\angle ZAC=\angle OAY=90^{\circ}</math>, <math>\angle ZAC+\angle BAC=\angle OAY+\angle BAC</math>, <math>\angle ZAB=\angle CAY</math>, and <math>AZ=AC</math>, <math>AB=AY</math>, <math>\triangle AZB \cong \triangle ACY</math> by <math>SAS</math>, <math>BZ=YC</math>. Because <math>AZ \perp AC</math>,  <math>\triangle ACY</math> is a <math>90^{\circ}</math> rotation about point <math>A</math> of <math>\triangle AZB</math>. So, <math>BZ \perp YC</math>.
 +
 +
Because <math>OZ</math> and <math>OY</math> is the radius of <math>\odot O</math>, <math>OZ=OY</math>. Because <math>O</math> is the midpoint of hypotenuse <math>AB</math>, <math>OB=OC</math>, <math>BZ=CY</math>, <math>\triangle OBZ \cong \triangle OCY</math> by <math>SSS</math>. Because <math>BZ \perp CY</math>,  <math>\triangle OCY</math> is a <math>90^{\circ}</math> rotation about point <math>O</math> of <math>\triangle OBZ</math>. So, <math>OB \perp OC</math>.
 +
 +
<math>\angle COB = 90^{\circ}</math>, <math>OC=OB</math>, <math>\triangle OCB</math> is isosceles right triangle, <math>\angle ABC=\angle OBC=45^{\circ}</math>. So, <math>\triangle ABC</math> is isosceles right triangle.
 +
 +
Therefore, <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>.
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=gDSIM9SAstk
 +
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Fileagingassisitant fileagingassisitant]
  
 
==See Also==
 
==See Also==

Latest revision as of 15:53, 2 November 2024

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 $W$ lie 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

Video Solution

https://www.youtube.com/watch?v=gDSIM9SAstk


~fileagingassisitant

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

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