Difference between revisions of "2021 AIME I Problems/Problem 11"

Revision as of 16:35, 29 May 2021

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)

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.

IN PROGRESS. NO EDIT PLEASE. A MILLION THANKS.

I WILL FINISH WITHIN ONE DAY.

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

@@ Line 32: / Line 32: @@
 ==Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Law of Cosines, Ptolemy's Theorem)==
+Suppose <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>E,</math> and let <math>\theta=\angle AEB.</math>
+By the <b>Converse of the Inscribed Angle Theorem</b>, if distinct points <math>X</math> and <math>Y</math> lie on the same side of <math>\overline{PQ}</math> (but not on <math>\overline{PQ}</math> itself) for which <math>\angle PXQ=\angle PYQ,</math> then <math>P,Q,X,</math> and <math>Y</math> are cyclic.
 <b>IN PROGRESS. NO EDIT PLEASE. A MILLION THANKS.
 I WILL FINISH WITHIN ONE DAY.</b>
-~MRENTHUSIASM
+~MRENTHUSIASM (inspired by Math Jams's <b>2021 AIME I Discussion</b>)
 ==See also==
 {{AIME box|year=2021|n=I|num-b=10|num-a=12}}
 {{MAA Notice}}

2021 AIME I (Problems • Answer Key • Resources)
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