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2021 AIME I Problems/Problem 11

Revision as of 18:07, 29 May 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Law of Cosines, Ptolemy's 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

File:2021 AIME I Problem 11 Diagram.png

~MRENTHUSIASM (by Geometry Expressions)

Solution 1

Leonard my dude's image.png

Let $O$ be the intersection of $AC$ and $BD$. Let $\theta = \angle AOB$.

Firstly, since $\angle AA_1D = \angle AD_1D = 90^\circ$, we deduce that $AA_1D_1D$ is cyclic. This implies that $\triangle A_1OD_1 \sim \triangle AOD$, with a ratio of $\frac{A_1O}{AO} = \cos \angle A_1OA = \cos \theta$. This means that $\frac{A_1D_1}{AD} = \cos \theta$. Similarly, $\frac{A_1B_1}{AB} = \frac{B_1C_1}{BC} = \frac{C_1D_1}{CD} = \cos \theta$. Hence \[A_1B_1 + B_1C_1 + C_1D_1 + D_1A_1 = (AB + BC + CD + DA)\cos \theta\] It therefore only remains to find $\cos \theta$.

From Ptolemy's theorem, we have that $(BD)(AC) = 4\times6+5\times7 = 59$. From Brahmagupta's Formula, $[ABCD] = \sqrt{(11-4)(11-5)(11-6)(11-7)} = 2\sqrt{210}$. But the area is also $\frac{1}{2}(BD)(AC)\sin\theta = \frac{59}{2}\sin\theta$, so $\sin \theta = \frac{4\sqrt{210}}{59} \implies \cos \theta = \frac{11}{59}$. Then the desired fraction is $(4+5+6+7)\cos\theta = \frac{242}{59}$ for an answer of $\boxed{301}$.

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, \[\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}\] 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 $\frac{24}{59}$ \[\cos{\theta}=2\cos^2{\frac{\theta}{2}}-1=\frac{\pm11}{59}\] In this context, $\cos{\theta}>0$. Thus, $\cos{\theta}=\frac{11}{59}$ \[Ans=22*\cos{\theta}=22*\frac{11}{59}=\frac{242}{59}=\frac{m}{n}\] \[m+n=242+59=\boxed{301}\] ~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 \[49 - x^2 = 16 - (\sqrt{59} - x)^2 \implies 92 = 2\sqrt{59}x \implies x = \frac{46}{\sqrt{59}} = \frac{46\sqrt{59}}{59}.\] Now let $y$ be $C_1D.$ (thus making $C_1B = \sqrt{59} - y$). Similarly, we have \[36 - y^2 = 25 - (\sqrt{59} - y)^2 \implies 70 = 2\sqrt{59}y \implies y = \frac{35}{\sqrt{59}} = \frac{35\sqrt{59}}{59}.\] 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 (Cyclic Quadrilaterals, Similar Triangles, Law of Cosines, Ptolemy's Theorem)

Suppose $\overline{AC}$ and $\overline{BD}$ intersect at $E,$ and let $\theta=\angle AEB.$

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, we conclude that quadrilaterals $ABA_1B_1,BCC_1B_1,CDC_1D_1,$ and $DAA_1D_1$ are all cyclic.

In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have $\angle EA_1B_1=\angle EAB$ and $\angle EB_1A_1=\angle EBA$ by angle chasing, from which $\triangle A_1B_1E \sim \triangle ABE$ by AA, with the ratio of similitude \[\frac{A_1B_1}{AB}=\underbrace{\frac{A_1E}{AE}}_{\substack{\text{right} \\ \triangle A_1AE}}=\underbrace{\frac{B_1E}{BE}}_{\substack{\text{right} \\ \triangle B_1BE}}=\cos\theta.\]

IN PROGRESS. NO EDIT PLEASE. A MILLION THANKS.

I WILL FINISH WITHIN ONE DAY.

~MRENTHUSIASM (inspired by Math Jams's 2021 AIME I Discussion)

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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