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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 15"

(added 'signature' to solution 2, (added solution 2 in previous edit, forgot to summarize it oops))
(→‎Solution 2: oops accidentally put \cong instead of \sim)
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==Solution 2==
 
==Solution 2==
  
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We have that <math>\triangle CRB \cong \triangle BAP.</math> Thus, <math>\frac{\overline{CB}}{\overline{CR}} = \frac{\overline{PB}}{\overline{AB}}</math>. Now, let the side length of the square be <math>s.</math> Then, by the Pythagorean theorem, <math>CR = \sqrt{x^2-36}.</math> Plugging all of this information in, we get <cmath>\frac{s}{\sqrt{s^2-36}} = \frac{13}{s}.</cmath> Simplifying gives <cmath>s^2=13\sqrt{s^2-36},</cmath> Squaring both sides gives <cmath>s^4 = 169s^2- 169\cdot 36 \implies s^4-169s^2 + 169\cdot 36 = 0.</cmath> We now set <math>s^2=t,</math> and get the equation <math>t^2-169t + 169\cdot 36 = 0.</math> From here, notice we want to solve for <math>t</math>, as it is precisely <math>s^2,</math> or the area of the square. So we use the [[Quadratic formula]], and though it may seem bashy, we hope for a nice cancellation of terms. <cmath>t = \frac{169\pm\sqrt{169^2-4\cdot 36 \cdot 169}}{2}.</cmath> It seems scary, but factoring <math>169</math> from the square root gives us <cmath>t = \frac{169\pm \sqrt{169 \cdot (169-144)}}{2} = \frac{169 \pm \sqrt{169 \cdot 25}}{2} = \frac{169 \pm 13\cdot 5}{2} = \frac{169\pm 65}{2},</cmath> giving us the solutions <cmath>t=52, 117.</cmath> We instantly see that <math>t=52</math> is way too small to be an area of this square (<math>52</math> isn't even an answer choice, so you can skip this step if out of time) because then the side length would be <math>2\sqrt{13}</math> and then, even the largest line you can draw inside the square (the diagonal)  is <math>2\sqrt{26},</math> which is less than <math>13</math> (line <math>PB</math>) And thus, <math>t</math> must be <math>117</math>, and our answer is <math>\boxed{\textbf{(D)}}.</math> <math>\blacksquare</math>
+
We have that <math>\triangle CRB \sim \triangle BAP.</math> Thus, <math>\frac{\overline{CB}}{\overline{CR}} = \frac{\overline{PB}}{\overline{AB}}</math>. Now, let the side length of the square be <math>s.</math> Then, by the Pythagorean theorem, <math>CR = \sqrt{x^2-36}.</math> Plugging all of this information in, we get <cmath>\frac{s}{\sqrt{s^2-36}} = \frac{13}{s}.</cmath> Simplifying gives <cmath>s^2=13\sqrt{s^2-36},</cmath> Squaring both sides gives <cmath>s^4 = 169s^2- 169\cdot 36 \implies s^4-169s^2 + 169\cdot 36 = 0.</cmath> We now set <math>s^2=t,</math> and get the equation <math>t^2-169t + 169\cdot 36 = 0.</math> From here, notice we want to solve for <math>t</math>, as it is precisely <math>s^2,</math> or the area of the square. So we use the [[Quadratic formula]], and though it may seem bashy, we hope for a nice cancellation of terms. <cmath>t = \frac{169\pm\sqrt{169^2-4\cdot 36 \cdot 169}}{2}.</cmath> It seems scary, but factoring <math>169</math> from the square root gives us <cmath>t = \frac{169\pm \sqrt{169 \cdot (169-144)}}{2} = \frac{169 \pm \sqrt{169 \cdot 25}}{2} = \frac{169 \pm 13\cdot 5}{2} = \frac{169\pm 65}{2},</cmath> giving us the solutions <cmath>t=52, 117.</cmath> We instantly see that <math>t=52</math> is way too small to be an area of this square (<math>52</math> isn't even an answer choice, so you can skip this step if out of time) because then the side length would be <math>2\sqrt{13}</math> and then, even the largest line you can draw inside the square (the diagonal)  is <math>2\sqrt{26},</math> which is less than <math>13</math> (line <math>PB</math>) And thus, <math>t</math> must be <math>117</math>, and our answer is <math>\boxed{\textbf{(D)}}.</math> <math>\blacksquare</math>
  
 
~wamofan
 
~wamofan

Revision as of 18:56, 24 November 2021

Problem

In square $ABCD$, points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$, respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$, with $BR = 6$ and $PR = 7$. What is the area of the square?

[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]

$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$

Solution

Note that $\triangle APB \cong \triangle BQC.$ Then, it follows that $\overline{PB} \cong \overline{QC}.$ Thus, $QC = PB = PR + RB = 7 + 6 = 13.$ Define $x$ to be the length of side $CR,$ then $RQ = 13-x.$ Because $\overline{BR}$ is the altitude of the triangle, we can use the property that $QR \cdot RC = BR^2.$ Substituting the given lengths, we have \[(13-x) \cdot x = 36.\] Solving, gives $x = 4$ and $x = 9.$ We eliminate the possibilty of $x=4$ because $RC > QR.$ Thus, the side lengnth of the square, by Pythagorean Theorem, is \[\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.\] Thus, the area of the sqaure is $(\sqrt{117})^2 = 117.$ Thus, the answer is $\boxed{(\textbf{D}.)}.$

~NH14

Solution 2

We have that $\triangle CRB \sim \triangle BAP.$ Thus, $\frac{\overline{CB}}{\overline{CR}} = \frac{\overline{PB}}{\overline{AB}}$. Now, let the side length of the square be $s.$ Then, by the Pythagorean theorem, $CR = \sqrt{x^2-36}.$ Plugging all of this information in, we get \[\frac{s}{\sqrt{s^2-36}} = \frac{13}{s}.\] Simplifying gives \[s^2=13\sqrt{s^2-36},\] Squaring both sides gives \[s^4 = 169s^2- 169\cdot 36 \implies s^4-169s^2 + 169\cdot 36 = 0.\] We now set $s^2=t,$ and get the equation $t^2-169t + 169\cdot 36 = 0.$ From here, notice we want to solve for $t$, as it is precisely $s^2,$ or the area of the square. So we use the Quadratic formula, and though it may seem bashy, we hope for a nice cancellation of terms. \[t = \frac{169\pm\sqrt{169^2-4\cdot 36 \cdot 169}}{2}.\] It seems scary, but factoring $169$ from the square root gives us \[t = \frac{169\pm \sqrt{169 \cdot (169-144)}}{2} = \frac{169 \pm \sqrt{169 \cdot 25}}{2} = \frac{169 \pm 13\cdot 5}{2} = \frac{169\pm 65}{2},\] giving us the solutions \[t=52, 117.\] We instantly see that $t=52$ is way too small to be an area of this square ($52$ isn't even an answer choice, so you can skip this step if out of time) because then the side length would be $2\sqrt{13}$ and then, even the largest line you can draw inside the square (the diagonal) is $2\sqrt{26},$ which is less than $13$ (line $PB$) And thus, $t$ must be $117$, and our answer is $\boxed{\textbf{(D)}}.$ $\blacksquare$

~wamofan

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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