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

Revision as of 09:19, 25 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

Solution 3

Tilt your head and treat $R, \overarrow{RB}$ (Error compiling LaTeX. Unknown error_msg) and $\overarrow{RC}$ (Error compiling LaTeX. Unknown error_msg) as the origin, $x$ -axis and $y$ -axis, respectively.

~VensL.

@@ Line 34: / Line 34: @@
 ~wamofan
+==Solution 3==
+Tilt your head and treat <math>R, \overarrow{RB}</math> and <math>\overarrow{RC}</math> as the origin, <math>x</math>-axis and <math>y</math>-axis, respectively.
+~VensL.
 ==See Also==
 {{AMC10 box|year=2021 Fall|ab=B|num-a=16|num-b=14}}
 {{MAA Notice}}

2021 Fall AMC 10B (Problems • Answer Key • Resources)
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