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

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<cmath>XR\cdot XA = XP\cdot XB,</cmath>which implies <math>\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34</math>.
 
<cmath>XR\cdot XA = XP\cdot XB,</cmath>which implies <math>\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34</math>.
  
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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}=15.</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{\textbf{(A)} ~81}</math>.
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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}=15</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{\textbf{(A)} ~81}</math>.
  
 
==Solution 2 (Trig) ==
 
==Solution 2 (Trig) ==
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<cmath> BD^2 = (2x+6)^2 = 4(x+3)^2 = 4a = 4 \cdot \frac{16 + \sqrt{656}}{2} = 32+8\sqrt{41}</cmath>  
 
<cmath> BD^2 = (2x+6)^2 = 4(x+3)^2 = 4a = 4 \cdot \frac{16 + \sqrt{656}}{2} = 32+8\sqrt{41}</cmath>  
 
and the desired answer is <math>32+8+41 = \boxed{\textbf{(A)} ~81}</math> ~skyscraper
 
and the desired answer is <math>32+8+41 = \boxed{\textbf{(A)} ~81}</math> ~skyscraper
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==Solution 6 (Similar Triangles & Pythagorean Theorem)==
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Let the intersection of <math>RS</math> and <math>BD</math> be <math>X</math>.
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<math>\because</math> <math>\angle APX = \angle DSX</math> and <math>\angle AXP = \angle DXS</math>, <math>\triangle APX \sim \triangle DSX</math> by <math>AA</math>
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<math>\therefore</math> <math>\frac{PA}{DS} = \frac68 = \frac34</math>, <math>DS = \frac43 \cdot PA</math>
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By the Pythagorean theorem and the property of projection, <math>BD^2 = (DS+BR)^2 + RS^2 = 4DS^2 + 64 = 4(\frac43 \cdot PA)^2 + 64 = \frac{64}{9} \cdot PA^2 + 64</math>, <math>\frac{64}{9} \cdot PA^2 = BD^2 - 64</math>
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<math>\because [ABCD] = PA \cdot BD = 15</math>, <math>\therefore PA = \frac{15}{BD}</math>
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<cmath>\frac{64}{9} (\frac{15}{BD})^2 = BD^2 - 64</cmath>
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<cmath>\frac{1600}{BD^2} = BD^2 - 64</cmath>
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<cmath>BD^4 - 64 BD^2 - 1600 = 0</cmath>
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<cmath>BD^2 = \frac{64 + \sqrt{64^2 - 4 (-1600)}}{2} = 32 + 8 \sqrt{41}</cmath>
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Therefore, the answer is <math>32 + 8 + 41 = \boxed{\textbf{(A) } 81}</math>.
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Solution 7==
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Let <math>O</math> be the intersection of the diagonals in quadrilateral <math>ABCD</math> and let the origin be at <math>O</math>.
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Then, let <math>BD</math> be on the x-axis, making <math>\vec{Q}=\begin{pmatrix}3 \\ 0\end{pmatrix}</math>, <math>\vec{C}=\begin{pmatrix}3 \\ y\end{pmatrix}</math>, <math>\vec{B}=\begin{pmatrix}x \\ 0\end{pmatrix}</math>, and <math>\vec{D}=\begin{pmatrix}-x \\ 0\end{pmatrix}</math>.
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The area of <math>\triangle BDC</math> is half the area of <math>ABCD</math>, so,
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<cmath>\frac{1}{2}(\overline{BD})(\overline{QC})=\frac{1}{2}(ABCD)</cmath>
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<cmath>\frac{1}{2}(2x)(y)=\frac{1}{2}(15)</cmath>
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<cmath>xy=\frac{15}{2}</cmath>
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<cmath>y= \frac{15}{2x}</cmath>.
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Also, <math>\vec{R}</math> is the projection of <math>\vec{B}</math> onto <math>\vec{AC}</math>, which is the same as projecting it onto <math>\vec{C}</math>. We get:
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<cmath>\vec{R}=proj_{\vec{C}}{(\vec{B})}=\frac{\vec{B} \cdot \vec{C}}{\vec{C} \cdot \vec{C}}(\vec{C})=\frac{\begin{pmatrix}x \\ 0\end{pmatrix} \cdot \begin{pmatrix}3 \\ y\end{pmatrix}}{\begin{pmatrix}3 \\ y\end{pmatrix} \cdot \begin{pmatrix}3 \\ y\end{pmatrix}}\begin{pmatrix}3 \\ y\end{pmatrix}=\frac{3x}{y^2+9}\begin{pmatrix}3 \\ y\end{pmatrix}</cmath>
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We are given that <math>\overline{RS} = 8</math>, so <math>||\vec{RS}|| = 8</math>. Because diagonals bisect each other in a parallelogram, <math>||\vec{R}|| = \frac{8}{2} = 4</math>. Substituting <math>||\vec{R}||</math> with the previous equation gives:
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<cmath>||\frac{3x}{y^2+9}\begin{pmatrix}3 \\ y\end{pmatrix}|| = 4</cmath>
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<cmath>\frac{3x}{y^2+9}\sqrt{y^2+9} = 4</cmath>
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<cmath>\frac{3x}{\sqrt{y^2+9}} = 4</cmath>
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Squaring both sides gives:
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<cmath>\frac{9x^2}{y^2+9}=16</cmath>
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<cmath>9x^2=16y^2+144</cmath>
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Substituting <math>y=\frac{15}{2x}</math>:
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<cmath>9x^2=16\left(\frac{15}{2x}\right)^2+144</cmath>
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<cmath>9x^2=16\left(\frac{225}{4x^2}\right)+144</cmath>
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<cmath>9x^4=144x^2+900</cmath>
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<cmath>x^4-16x^2-100=0</cmath>
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Solving the quadratic for <math>x^2</math> gives:
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<cmath>x^2 = \frac{16 \pm \sqrt{16^2 - 4(1)(-100)}}{2} = 8 \pm 2\sqrt{41}</cmath>
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But <math>x^2</math> is nonnegative, so <math>x^2</math> must be <math>8 + 2\sqrt{41}</math>. We are looking for <math>\overline{BD}^2</math>, which is <math>4x^2</math>. <math>4(8 + 2\sqrt{41}) = 32 + 8\sqrt{41}</math>, so <math>a + b + c = 32 + 8 + 41 = \boxed{\textbf{(A) } 81}</math> ~askeww (My first solution ever!)
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==Video Solution by MOP 2024==
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https://youtube.com/watch?v=xtYSPxOMZlk
  
 
== Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point) ==
 
== Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point) ==

Latest revision as of 20:14, 28 May 2024

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\[[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}=15\]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 $32 + 8 + 41 = \boxed{\textbf{(A)} ~81}$.

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 $d^2 = 64x = 32 + 8\sqrt{41} \rightarrow 32+8+41=\boxed{\textbf{(A)} ~81}$

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 $d^2 = \frac{64}{9}(x^2+9) = 8\sqrt{41} + 32 \Rightarrow 8+41+32=\boxed{\textbf{(A)} ~81}$

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 $d^2=\frac{64+\sqrt{4096+6400}}{2}=32+8\sqrt{41} \rightarrow 32+8+41= \boxed{\textbf{(A)} ~81}$

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 $32+8+41 = \boxed{\textbf{(A)} ~81}$ ~skyscraper

Solution 6 (Similar Triangles & Pythagorean Theorem)

Let the intersection of $RS$ and $BD$ be $X$.

$\because$ $\angle APX = \angle DSX$ and $\angle AXP = \angle DXS$, $\triangle APX \sim \triangle DSX$ by $AA$

$\therefore$ $\frac{PA}{DS} = \frac68 = \frac34$, $DS = \frac43 \cdot PA$

By the Pythagorean theorem and the property of projection, $BD^2 = (DS+BR)^2 + RS^2 = 4DS^2 + 64 = 4(\frac43 \cdot PA)^2 + 64 = \frac{64}{9} \cdot PA^2 + 64$, $\frac{64}{9} \cdot PA^2 = BD^2 - 64$

$\because [ABCD] = PA \cdot BD = 15$, $\therefore PA = \frac{15}{BD}$

\[\frac{64}{9} (\frac{15}{BD})^2 = BD^2 - 64\]

\[\frac{1600}{BD^2} = BD^2 - 64\]

\[BD^4 - 64 BD^2 - 1600 = 0\]

\[BD^2 = \frac{64 + \sqrt{64^2 - 4 (-1600)}}{2} = 32 + 8 \sqrt{41}\]

Therefore, the answer is $32 + 8 + 41 = \boxed{\textbf{(A) } 81}$.

~isabelchen

Solution 7

Let $O$ be the intersection of the diagonals in quadrilateral $ABCD$ and let the origin be at $O$.

Then, let $BD$ be on the x-axis, making $\vec{Q}=\begin{pmatrix}3 \\ 0\end{pmatrix}$, $\vec{C}=\begin{pmatrix}3 \\ y\end{pmatrix}$, $\vec{B}=\begin{pmatrix}x \\ 0\end{pmatrix}$, and $\vec{D}=\begin{pmatrix}-x \\ 0\end{pmatrix}$.

The area of $\triangle BDC$ is half the area of $ABCD$, so, \[\frac{1}{2}(\overline{BD})(\overline{QC})=\frac{1}{2}(ABCD)\] \[\frac{1}{2}(2x)(y)=\frac{1}{2}(15)\] \[xy=\frac{15}{2}\] \[y= \frac{15}{2x}\].

Also, $\vec{R}$ is the projection of $\vec{B}$ onto $\vec{AC}$, which is the same as projecting it onto $\vec{C}$. We get: \[\vec{R}=proj_{\vec{C}}{(\vec{B})}=\frac{\vec{B} \cdot \vec{C}}{\vec{C} \cdot \vec{C}}(\vec{C})=\frac{\begin{pmatrix}x \\ 0\end{pmatrix} \cdot \begin{pmatrix}3 \\ y\end{pmatrix}}{\begin{pmatrix}3 \\ y\end{pmatrix} \cdot \begin{pmatrix}3 \\ y\end{pmatrix}}\begin{pmatrix}3 \\ y\end{pmatrix}=\frac{3x}{y^2+9}\begin{pmatrix}3 \\ y\end{pmatrix}\]

We are given that $\overline{RS} = 8$, so $||\vec{RS}|| = 8$. Because diagonals bisect each other in a parallelogram, $||\vec{R}|| = \frac{8}{2} = 4$. Substituting $||\vec{R}||$ with the previous equation gives: \[||\frac{3x}{y^2+9}\begin{pmatrix}3 \\ y\end{pmatrix}|| = 4\] \[\frac{3x}{y^2+9}\sqrt{y^2+9} = 4\] \[\frac{3x}{\sqrt{y^2+9}} = 4\] Squaring both sides gives: \[\frac{9x^2}{y^2+9}=16\] \[9x^2=16y^2+144\] Substituting $y=\frac{15}{2x}$: \[9x^2=16\left(\frac{15}{2x}\right)^2+144\] \[9x^2=16\left(\frac{225}{4x^2}\right)+144\] \[9x^4=144x^2+900\] \[x^4-16x^2-100=0\] Solving the quadratic for $x^2$ gives: \[x^2 = \frac{16 \pm \sqrt{16^2 - 4(1)(-100)}}{2} = 8 \pm 2\sqrt{41}\] But $x^2$ is nonnegative, so $x^2$ must be $8 + 2\sqrt{41}$. We are looking for $\overline{BD}^2$, which is $4x^2$. $4(8 + 2\sqrt{41}) = 32 + 8\sqrt{41}$, so $a + b + c = 32 + 8 + 41 = \boxed{\textbf{(A) } 81}$ ~askeww (My first solution ever!)

Video Solution by MOP 2024

https://youtube.com/watch?v=xtYSPxOMZlk

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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