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Difference between revisions of "2021 AMC 12B Problems/Problem 24"

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==Solution 3 (Similar Triangles and Algebra)==
 
==Solution 3 (Similar Triangles and Algebra)==
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Let <math>X</math> be the intersection of diagonals <math>AC</math> and <math>BD</math>. By symmetry <math>[\triangle XCB] = \frac{15}{4}</math>, <math>XQ = 3</math> and <math>XR = 4</math>, so now we have reduced all of the conditions one quadrant. Let <math>CQ = x</math>. <math>XC = \sqrt{x^2+9}</math>, <math>RB = \frac{4x}{3}</math> by similar triangles and using the area condition we get <math>\frac{4}{3} \cdot x \cdot \sqrt{x^2+9} = \frac{15}{2}</math>. Note that it suffices to find <math>XB = \frac{4}{3}\sqrt{x^2+9}</math> because we can double and square it to get <math>d^2</math>. Solving for <math>a = x^2</math> in the above equation, and then using <math>d^2 = \frac{64}{9}(x^2+9) = 8\sqrt{41} + 32 \Rightarrow \boxed{81}</math>. (edited by vaporwave)
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Let <math>X</math> be the intersection of diagonals <math>AC</math> and <math>BD</math>. By symmetry <math>[\triangle XCB] = \frac{15}{4}</math>, <math>XQ = 3</math> and <math>XR = 4</math>, so now we have reduced all of the conditions one quadrant. Let <math>CQ = x</math>. <math>XC = \sqrt{x^2+9}</math>, <math>RB = \frac{4x}{3}</math> by similar triangles and using the area condition we get <math>\frac{4}{3} \cdot x \cdot \sqrt{x^2+9} = \frac{15}{2}</math>. Note that it suffices to find <math>XB = \frac{4}{3}\sqrt{x^2+9}</math> because we can double and square it to get <math>d^2</math>. Solving for <math>a = x^2</math> in the above equation, and then using <math>d^2 = \frac{64}{9}(x^2+9) = 8\sqrt{41} + 32 \Rightarrow \boxed{81}</math>.
  
 
==Solution 4 (Similar Triangles)==
 
==Solution 4 (Similar Triangles)==

Revision as of 17:46, 20 August 2022

Problem

Let $ABCD$ be a parallelogram with area $15$. Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.

[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]

Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$

$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$

Solution 1

Let $X$ denote the intersection point of the diagonals $AC$ and $BD$. Remark that by symmetry $X$ is the midpoint of both $\overline{PQ}$ and $\overline{RS}$, so $XP = XQ = 3$ and $XR = XS = 4$. Now note that since $\angle APB = \angle ARB = 90^\circ$, quadrilateral $ARPB$ is cyclic, and so \[XR\cdot XA = XP\cdot XB,\]which implies $\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34$.

Thus let $x> 0$ be such that $XA = 3x$ and $XB = 4x$. Then Pythagorean Theorem on $\triangle APX$ yields $AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}$, and so\[[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.\]Solving this for $x^2$ yields $x^2 = \tfrac12 + \tfrac{\sqrt{41}}8$, and so\[(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.\]The requested answer is $32 + 8 + 41 = \boxed{81}$.

Solution 2 (Trig)

Let $X$ denote the intersection point of the diagonals $AC$ and $BD,$ and let $\theta = \angle{COB}$. Then, by the given conditions, $XR = 4,$ $XQ = 3,$ $[XCB] = \frac{15}{4}$. So, \[XC = \frac{3}{\cos \theta}\] \[XB \cos \theta = 4\] \[\frac{1}{2} XB XC \sin \theta = \frac{15}{4}\] Combining the above 3 equations, we get \[\frac{\sin \theta }{\cos^2 \theta} = \frac{5}{8}.\] Since we want to find $d^2 = 4XB^2 = \frac{64}{\cos^2 \theta},$ we let $x = \frac{1}{\cos^2 \theta}.$ Then \[\frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}.\] Solving this, we get $x = \frac{4 + \sqrt{41}}{8},$ so $d^2 = 64x = 32 + 8\sqrt{41}$. $\boxed{81}$

Solution 3 (Similar Triangles and Algebra)

Let $X$ be the intersection of diagonals $AC$ and $BD$. By symmetry $[\triangle XCB] = \frac{15}{4}$, $XQ = 3$ and $XR = 4$, so now we have reduced all of the conditions one quadrant. Let $CQ = x$. $XC = \sqrt{x^2+9}$, $RB = \frac{4x}{3}$ by similar triangles and using the area condition we get $\frac{4}{3} \cdot x \cdot \sqrt{x^2+9} = \frac{15}{2}$. Note that it suffices to find $XB = \frac{4}{3}\sqrt{x^2+9}$ because we can double and square it to get $d^2$. Solving for $a = x^2$ in the above equation, and then using $d^2 = \frac{64}{9}(x^2+9) = 8\sqrt{41} + 32 \Rightarrow \boxed{81}$.

Solution 4 (Similar Triangles)

Again, Let $X$ be the intersection of diagonals $AC$ and $BD$. Note that triangles $\triangle QXC$ and $\triangle BXR$ are similar because they are right triangles and share $\angle CXQ$. First, call the length of $XB = \frac{d}{2}$. By the definition of an area of a parallelogram, $CQ \cdot 2XB = 15$, so $CQ = \frac{15}{d}$. Using similar triangles on $\triangle QXC$ and $\triangle BXR$, $\frac{CQ}{XQ} = \frac{BR}{XR}$. Therefore, finding $BR$, $BR = \frac{XR}{XQ} \cdot CQ = \frac{4}{3} \cdot \frac{15}{d} = \frac{20}{d}$. Now, applying the Pythagorean theorem once, we find $(\frac{20}{d}) ^2$ + $(4)^2$ = $(\frac{d}{2}) ^2$. Solving this equation for $d^2$, we find $d^2=\frac{64+\sqrt{4096+6400}}{2}=32+8\sqrt{41} \Rightarrow \boxed{81}$.

Solution 5

Let $BQ = PD = x.$ We know that the area of the parallelogram is $15,$ so it follows that $[\triangle{BCD}] = [\triangle{BAD}] = \tfrac{15}{2}$ and the height of each triangle, which are also the lengths of $QC$ and $AP,$ is $\tfrac{15}{2(x+3)}.$ Suppose that $E = RS \cap BD.$ Because $\angle{BRE} = \angle{CQE}$ and $\angle{BER} = \angle{CQD},$ we have $\triangle{BRE} \sim \triangle{CQE}.$ The length of $CE,$ by the Pythagorean Theorem is $\sqrt{3^2+(\tfrac{15}{2(x+3)})^2}$ and the length of $BR,$ by the Pythagorean Theorem on $\triangle{BRE},$ is $\sqrt{(x+3)^2 - 4}.$ Note that \[\sin{\angle QEC} = \frac{CQ}{CE} = \frac{BR}{BE}\] Substituting in our values, \[\frac{\frac{15}{2(x+3)}}{\sqrt{9+(\frac{15}{2(x+3)})^2}} = \frac{\sqrt{(x+3)^2 - 4^2}}{x+3}\] To rid unnecessary computation, we let $(x+3)^2 = a.$ The equation simplifies, after cross multiplying, to \[\sqrt{9+\frac{15^2}{4a}} \sqrt{a-16 } = \frac{15}{2}\] \[36a^2 - 576a - 15^2\cdot 16 = 0\] \[a^2-16a-100 =0\] By the quadratic formula, $a \in \{\tfrac{16 - \sqrt{656}}{2}, \tfrac{16 + \sqrt{656}}{2}\},$ so we discard the negative solution. The value of $BD^2$ is \[BD^2 = (2x+6)^2 = 4(x+3)^2 = 4a = 4 \cdot \frac{16 + \sqrt{656}}{2} = 32+8\sqrt{41}\] and the desired answer is $32+8+41 = 81.$ ~skyscraper

Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point)

https://youtu.be/1zhwR9B2Gy8

~ pi_is_3.14

Video Solution (Simple: Using trigonometry and Equations)

https://youtu.be/ZB-VN02H6mU ~hippopotamus1

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 12 Problems and Solutions

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