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

(Solution 2)
(Solution 2)
 
(10 intermediate revisions by 3 users not shown)
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==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 = \angle WXB</math>. You can use the same method to find that <math>\angle{ZYA}</math> = <math>\angle{XWB}</math>. 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>  
  
<asy>  
+
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>.  
/* 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 */
 
  
 +
That means <math>\triangle ZYA \sim \triangle XWB</math>.
  
draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle);
+
Setting up proportions,  
draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle);
+
<math>\dfrac{b}{12}=\dfrac{12}{a+b}.</math>
/* draw figures */
+
Cross-multiplying we get:
draw((3.46,0.96)--(3.44,-3.36));
+
<math>b^2+ab=12^2</math>
draw((3.44,-3.36)--(8.02,-3.44));
+
 
draw((8.02,-3.44)--(3.46,0.96));
+
But also, by Pythagoras,  
draw((3.46,0.96)--(-0.86,0.98));
+
<math>b^2+a^2=12^2</math>, so <math>ab=a^2 \Rightarrow a=b</math>
draw((-0.86,0.98)--(-0.88,-3.34));
+
 
draw((-0.88,-3.34)--(3.44,-3.36));
+
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 <cmath>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</cmath>
draw((3.46,0.96)--(8.02,-3.44));
+
 
draw((8.02,-3.44)--(12.42,1.12));
+
~BakedPotato66
draw((12.42,1.12)--(7.86,5.52));
+
 
draw((7.86,5.52)--(3.46,0.96));
+
~LegionOfAvatars
draw((5.74,-1.24)--(-0.86,0.98));
+
 
draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4"));
+
==Solution 3 - Fakesolve==
draw((5.74,-1.24)--(7.86,5.52));
+
Temporarily assume that we are as dumb as samrocksnature. Then, we would misread the instructions and draw the squares inside, not outside, the triangle. In addition, we would make a plethora of ridiculous assumptions.
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"));
+
Our diagram would essentially be two squares intersecting in a square, such that a circle can be drawn through the two leftmost vertices of the left square and the two rightmost vertices of the right square. Since the diagonal of the square is given to be <math>12</math>, our answer is <math>12+6\sqrt{2}+6\sqrt{2} \Rightarrow \boxed{C}</math>.
/* 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==
 
==See Also==

Latest revision as of 01:34, 20 January 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 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 \[\boxed{\textbf{(C)}\ 12+12\sqrt{2}}\]

~BakedPotato66

~LegionOfAvatars

Solution 3 - Fakesolve

Temporarily assume that we are as dumb as samrocksnature. Then, we would misread the instructions and draw the squares inside, not outside, the triangle. In addition, we would make a plethora of ridiculous assumptions.

Our diagram would essentially be two squares intersecting in a square, such that a circle can be drawn through the two leftmost vertices of the left square and the two rightmost vertices of the right square. Since the diagonal of the square is given to be $12$, our answer is $12+6\sqrt{2}+6\sqrt{2} \Rightarrow \boxed{C}$.

See Also

2015 AMC 10B (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 10 Problems and Solutions

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