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

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Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math>. Remark that by symmetry <math>X</math> is the midpoint of both <math>\overline{PQ}</math> and <math>\overline{RS}</math>, so <math>XP = XQ = 3</math> and <math>XR = XS = 4</math>. Now note that since <math>\angle APB = \angle ARB = 90^\circ</math>, quadrilateral <math>ARPB</math> is cyclic, and so
 
Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math>. Remark that by symmetry <math>X</math> is the midpoint of both <math>\overline{PQ}</math> and <math>\overline{RS}</math>, so <math>XP = XQ = 3</math> and <math>XR = XS = 4</math>. Now note that since <math>\angle APB = \angle ARB = 90^\circ</math>, quadrilateral <math>ARPB</math> is cyclic, and so
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<cmath>XR\cdot XA = XP\cdot XB,</cmath>which implies <math>\tfrac{XB}{XA} = \tfrac{XP}{XR} = \tfrac34</math>.
+
<cmath>XR\cdot XA = XP\cdot XB,</cmath>which implies <math>\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34</math>.
  
 
Thus let <math>x> 0</math> be such that <math>XA = 3x</math> and <math>XB = 4x</math>. Then Pythagorean Theorem on <math>\triangle APX</math> yields <math>AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}</math>, and so<cmath>[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.</cmath>Solving this for <math>x^2</math> yields <math>x^2 = \tfrac12 + \tfrac{\sqrt{41}}8</math>, and so<cmath>(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.</cmath>The requested answer is <math>32 + 8 + 41 = \boxed{81}</math>.
 
Thus let <math>x> 0</math> be such that <math>XA = 3x</math> and <math>XB = 4x</math>. Then Pythagorean Theorem on <math>\triangle APX</math> yields <math>AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}</math>, and so<cmath>[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.</cmath>Solving this for <math>x^2</math> yields <math>x^2 = \tfrac12 + \tfrac{\sqrt{41}}8</math>, and so<cmath>(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.</cmath>The requested answer is <math>32 + 8 + 41 = \boxed{81}</math>.

Revision as of 21:37, 15 February 2021

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 $m+n+p?$

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

simple Video Solution Using trigonometry and Equations

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

Solution

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, \[XC = \frac{3}{\cos \theta}\] \[XB \cos \theta = 4\] \[\frac{1}{2} XB XC \sin \theta = \frac{15}{4}\] Combine the above 3 equations we get \[\frac{\sin \theta }{\cos^2 \theta} = \frac{5}{8}\]. Since we want to find $d^2 = 4XB^2 = \frac{64}{\cos^2 \theta}$, let $x = \frac{1}{\cos^2 \theta}$, then \[\frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}\]. Solve, we get $x = \frac{4 + \sqrt{41}}{8}$, so $d^2 = 64x = 32 + 8\sqrt{41}$. $\boxed{81}$

~sequoia

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

https://youtu.be/1zhwR9B2Gy8

~ pi_is_3.14

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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