2021 AIME I Problems/Problem 11

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

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)

This solution refers to the Diagram section.

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 $\frac{A_1B_1}{AB}=\underbrace{\frac{A_1E}{AE}}_{\substack{\text{right} \\ \triangle A_1AE}}=\underbrace{\frac{B_1E}{BE}}_{\substack{\text{right} \\ \triangle B_1BE}}=\cos\theta. \hspace{15mm}(1)$ Similarly, we have $\angle EC_1D_1=\angle ECD$ and $\angle ED_1C_1=\angle EDC$ by angle chasing, 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 $(2):$

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

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

IN PROGRESS. NO EDIT PLEASE. A MILLION THANKS.

I WILL FINISH WITHIN ONE DAY.

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

2021 AIME I (Problems • Answer Key • Resources)
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