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2021 AIME I Problems/Problem 11

Revision as of 17:27, 28 May 2022 by Vvsss (talk | contribs) (→‎Solution 3 (Pythagorean Theorem))

Problem

Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Diagram

[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1; A = origin; C = (sqrt(53041)/31,0); B = intersectionpoints(Circle(A,4),Circle(C,5))[0]; D = intersectionpoints(Circle(A,7),Circle(C,6))[1]; A1 = foot(A,B,D); C1 = foot(C,B,D); B1 = foot(B,A,C); D1 = foot(D,A,C); markscalefactor=0.025; draw(rightanglemark(A,A1,B),red); draw(rightanglemark(B,B1,A),red); draw(rightanglemark(C,C1,D),red); draw(rightanglemark(D,D1,C),red); draw(A1--B1--C1--D1--cycle,green); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*dir(180-aCos(11/59)),linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$D$",D,1.5*dir(-aCos(11/59)),linewidth(4)); dot("$A_1$",A1,1.5*dir(A1-A),linewidth(4)); dot("$B_1$",B1,1.5*S,linewidth(4)); dot("$C_1$",C1,1.5*dir(C1-C),linewidth(4)); dot("$D_1$",D1,1.5*N,linewidth(4)); draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C)); draw(A--A1^^B--B1^^C--C1^^D--D1,dashed); [/asy] ~MRENTHUSIASM

Solution 1 (Cyclic Quadrilaterals, Similar Triangles, and Trigonometry)

This solution refers to the Diagram section.

By the Converse of the Inscribed Angle Theorem, if distinct points $X$ and $Y$ lie on the same side of $\overline{PQ}$ (but not on $\overline{PQ}$ itself) for which $\angle PXQ=\angle PYQ,$ then $P,Q,X,$ and $Y$ are cyclic. From the Converse of the Inscribed Angle Theorem, quadrilaterals $ABA_1B_1,BCC_1B_1,CDC_1D_1,$ and $DAA_1D_1$ are all cyclic.

Suppose $\overline{AC}$ and $\overline{BD}$ intersect at $E,$ and let $\angle AEB=\theta.$ It follows that $\angle CED=\theta$ and $\angle BEC=\angle DEA=180^\circ-\theta.$

We obtain the following diagram: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, M1, M2; A = origin; C = (sqrt(53041)/31,0); B = intersectionpoints(Circle(A,4),Circle(C,5))[0]; D = intersectionpoints(Circle(A,7),Circle(C,6))[1]; A1 = foot(A,B,D); C1 = foot(C,B,D); B1 = foot(B,A,C); D1 = foot(D,A,C); P = intersectionpoint(A--C,B--D); M1 = midpoint(A--B); M2 = midpoint(C--D); markscalefactor=0.025; draw(rightanglemark(A,A1,B),red); draw(rightanglemark(B,B1,A),red); draw(rightanglemark(C,C1,D),red); draw(rightanglemark(D,D1,C),red); draw(Arc(M1,A,B)^^Arc(M2,C,D),blue); draw(A1--B1--C1--D1--cycle,green); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*dir(180-aCos(11/59)),linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$D$",D,1.5*dir(-aCos(11/59)),linewidth(4)); dot("$A_1$",A1,1.5*dir(A1-A),linewidth(4)); dot("$B_1$",B1,1.5*S,linewidth(4)); dot("$C_1$",C1,1.5*dir(C1-C),linewidth(4)); dot("$D_1$",D1,1.5*N,linewidth(4)); dot("$E$",P,dir((180-aCos(11/59))/2),linewidth(4)); label("$\theta$",P,dir(180-aCos(11/59)/2),red); draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C)); draw(A--A1^^B--B1^^C--C1^^D--D1,dashed); [/asy] In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have $\angle EA_1B_1=\angle EAB$ (both supplementary to $\angle B_1A_1B$) and $\angle EB_1A_1=\angle EBA$ (both supplementary to $\angle A_1B_1A$), from which $\triangle A_1B_1E \sim \triangle ABE$ by AA, with the ratio of similitude \[\frac{A_1B_1}{AB}=\underbrace{\frac{A_1E}{AE}}_{\substack{\text{right} \\ \triangle A_1AE}}=\underbrace{\frac{B_1E}{BE}}_{\substack{\text{right} \\ \triangle B_1BE}}=\cos\theta. \hspace{15mm}(1)\] Similarly, we have $\angle EC_1D_1=\angle ECD$ (both supplementary to $\angle D_1C_1D$) and $\angle ED_1C_1=\angle EDC$ (both supplementary to $\angle C_1D_1C$), from which $\triangle C_1D_1E \sim \triangle CDE$ by AA, with the ratio of similitude \[\frac{C_1D_1}{CD}=\underbrace{\frac{C_1E}{CE}}_{\substack{\text{right} \\ \triangle C_1CE}}=\underbrace{\frac{D_1E}{DE}}_{\substack{\text{right} \\ \triangle D_1DE}}=\cos\theta. \hspace{14.75mm}(2)\] We apply the Transitive Property to $(1)$ and $(2):$

  1. We get $\frac{B_1E}{BE}=\frac{C_1E}{CE}=\cos\theta,$ so $\triangle B_1C_1E \sim \triangle BCE$ by SAS, with the ratio of similitude \[\frac{B_1C_1}{BC}=\frac{B_1E}{BE}=\frac{C_1E}{CE}=\cos\theta. \hspace{14.75mm}(3)\]

  2. We get $\frac{D_1E}{DE}=\frac{A_1E}{AE}=\cos\theta,$ so $\triangle D_1A_1E \sim \triangle DAE$ by SAS, with the ratio of similitude \[\frac{D_1A_1}{DA}=\frac{D_1E}{DE}=\frac{A_1E}{AE}=\cos\theta. \hspace{14mm}(4)\]

From $(1),(2),(3),$ and $(4),$ the perimeter of $A_1B_1C_1D_1$ is \begin{align*} A_1B_1+B_1C_1+C_1D_1+D_1A_1&=AB\cos\theta+BC\cos\theta+CD\cos\theta+DA\cos\theta \\ &=(AB+BC+CD+DA)\cos\theta \\ &=22\cos\theta. &&\hspace{5mm}(\bigstar) \end{align*} Two solutions follow from here:

Solution 1.1 (Law of Cosines)

Note that $\cos(180^\circ-\theta)=-\cos\theta$ holds for all $\theta.$ We apply the Law of Cosines to $\triangle ABE, \triangle BCE, \triangle CDE,$ and $\triangle DAE,$ respectively: \begin{alignat*}{12} &&&AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\angle AEB&&=AB^2&&\quad\implies\quad AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\theta&&=4^2, \hspace{15mm} &(1\star) \\ &&&BE^2+CE^2-2\cdot BE\cdot CE\cdot\cos\angle BEC&&=BC^2&&\quad\implies\quad BE^2+CE^2+2\cdot BE\cdot CE\cdot\cos\theta&&=5^2, \hspace{15mm} &(2\star) \\ &&&CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\angle CED&&=CD^2&&\quad\implies\quad CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\theta&&=6^2, \hspace{15mm} &(3\star) \\ &&&DE^2+AE^2-2\cdot DE\cdot AE\cdot\cos\angle DEA&&=DA^2&&\quad\implies\quad DE^2+AE^2+2\cdot DE\cdot AE\cdot\cos\theta&&=7^2. \hspace{15mm} &(4\star) \\ \end{alignat*} We subtract $(1\star)+(3\star)$ from $(2\star)+(4\star):$ \begin{align*} 2\cdot AE\cdot BE\cdot\cos\theta+2\cdot BE\cdot CE\cdot\cos\theta+2\cdot CE\cdot DE\cdot\cos\theta+2\cdot DE\cdot AE\cdot\cos\theta&=22 \\ 2\cdot\cos\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ })&=22 \\ 2\cdot\cos\theta\cdot59&=22 \\ \cos\theta&=\frac{11}{59}. \end{align*} Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$

~MRENTHUSIASM (credit given to Math Jams's 2021 AIME I Discussion)

Solution 1.2 (Area Formulas)

Let the brackets denote areas. We find $[ABCD]$ in two different ways:

  1. Note that $\sin(180^\circ-\theta)=\sin\theta$ holds for all $\theta.$ By area addition, we get \begin{align*} [ABCD]&=[ABE]+[BCE]+[CDE]+[DAE] \\ &=\frac12\cdot AE\cdot BE\cdot\sin\angle AEB+\frac12\cdot BE\cdot CE\cdot\sin\angle BEC+\frac12\cdot CE\cdot DE\cdot\sin\angle CED+\frac12\cdot DE\cdot AE\cdot\sin\angle DEA \\ &=\frac12\cdot AE\cdot BE\cdot\sin\theta+\frac12\cdot BE\cdot CE\cdot\sin\theta+\frac12\cdot CE\cdot DE\cdot\sin\theta+\frac12\cdot DE\cdot AE\cdot\sin\theta \\ &=\frac12\cdot\sin\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ }) \\ &=\frac12\cdot\sin\theta\cdot59. \end{align*}

  2. By Brahmagupta's Formula, we get \[[ABCD]=\sqrt{(s-AB)(s-BC)(s-CD)(s-DA)}=2\sqrt{210},\] where $s=\frac{AB+BC+CD+DA}{2}=11$ is the semiperimeter of $ABCD.$

Equating the expressions for $[ABCD],$ we have \[\frac12\cdot\sin\theta\cdot59=2\sqrt{210},\] so $\sin\theta=\frac{4\sqrt{210}}{59}.$ Since $0^\circ<\theta<90^\circ,$ we have $\cos\theta>0.$ It follows that \[\cos\theta=\sqrt{1-\sin^2\theta}=\frac{11}{59}.\] Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$

~MRENTHUSIASM (credit given to Leonard my dude)

Remark (Ptolemy's Theorem)

In $ABCD,$ we have \begin{align*} AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE &= (AE+CE)(BE+DE) &&\hspace{10mm}\text{Factor by Grouping} \\ &=AC\cdot BD &&\hspace{10mm}\text{Segment Addition} \\ &=AB\cdot CD+BC\cdot DA &&\hspace{10mm}\text{Ptolemy's Theorem} \\ &=59. &&\hspace{10mm}\text{Substitution} \end{align*} ~MRENTHUSIASM

Solution 2 (Finding cos x)

The angle $\theta$ between diagonals satisfies \[\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}\] (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). Thus, \[\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}\text{ or }\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4)(11-6)}}.\] That is, $\tan^2{\frac{\theta}{2}}=\frac{1-\cos^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}}=\frac{24}{35}$ or $\frac{35}{24}$. Thus, $\cos^2{\frac{\theta}{2}}=\frac{35}{59}$ or $\frac{24}{59}$. So, \[\cos{\theta}=2\cos^2{\frac{\theta}{2}}-1=\pm\frac{11}{59}.\] In this context, $\cos{\theta}>0$. Thus, $\cos{\theta}=\frac{11}{59}$. The perimeter of $A_1B_1C_1D_1$ is \[22\cdot\cos{\theta}=22\cdot\frac{11}{59}=\frac{242}{59},\] and the answer is $m+n=242+59=\boxed{301}$.

~y.grace.yu

Solution 3 (Pythagorean Theorem)

We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length $\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.$ [I don't believe this is correct... are the two diagonals of $ABCD$ necessarily congruent? -peace09] WLOG we focus on diagonal $BD.$ To find the diagonal of the inner quadrilateral, we drop the altitude from $A$ and $C$ and calculate the length of $A_1C_1.$ Let $x$ be $A_1D$ (Thus $A_1B = \sqrt{59} - x.$ By Pythagorean theorem, we have \[49 - x^2 = 16 - \left(\sqrt{59} - x\right)^2 \implies 92 = 2\sqrt{59}x \implies x = \frac{46}{\sqrt{59}} = \frac{46\sqrt{59}}{59}.\] Now let $y$ be $C_1D.$ (thus making $C_1B = \sqrt{59} - y$). Similarly, we have \[36 - y^2 = 25 - \left(\sqrt{59} - y\right)^2 \implies 70 = 2\sqrt{59}y \implies y = \frac{35}{\sqrt{59}} = \frac{35\sqrt{59}}{59}.\] We see that $A_1C_1$, the scaled down diagonal is just $x - y = \frac{11\sqrt{59}}{59},$ which is $\frac{\frac{11\sqrt{59}}{59}}{\sqrt{59}} = \frac{11}{59}$ times our original diagonal $BD,$ implying a scale factor of $\frac{11}{59}.$ Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply $\frac{11}{59} \cdot 22 = \frac{242}{59},$ making our answer $242+59 = \boxed{301}.$

~fidgetboss_4000


Solution 4 (Symmetry)

Lemma 1

In the triangle $ABC$, the points $B'$ and $C'$ are the bases of the heights dropped from the vertices $B$ and $C$, respectively. $\angle A = \alpha$. Then $B'C' = BC \cos\alpha,$ $\alpha < \frac{\pi}{2}$ and $B'C' = BC \cos (\pi – \alpha), \alpha >\frac{\pi}{2}.$

See Also

2021 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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