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2021 AIME I Problems/Problem 11

Revision as of 01:30, 12 March 2021 by Y.grace.yu (talk | contribs) (→‎Added Alternative Solution)

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 $m+n$.

Solution

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 $\boxed{301}$.

Finding $\cos{x}$ 2

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

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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