Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (Made the diagram more stand out. All solutions can refer to it.)
Line 2: Line 2:
 
Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB=4,BC=5,CD=6,</math> and <math>DA=7</math>. Let <math>A_1</math> and <math>C_1</math> be the feet of the perpendiculars from <math>A</math> and <math>C</math>, respectively, to line <math>BD,</math> and let <math>B_1</math> and <math>D_1</math> be the feet of the perpendiculars from <math>B</math> and <math>D,</math> respectively, to line <math>AC</math>. The perimeter of <math>A_1B_1C_1D_1</math> is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB=4,BC=5,CD=6,</math> and <math>DA=7</math>. Let <math>A_1</math> and <math>C_1</math> be the feet of the perpendiculars from <math>A</math> and <math>C</math>, respectively, to line <math>BD,</math> and let <math>B_1</math> and <math>D_1</math> be the feet of the perpendiculars from <math>B</math> and <math>D,</math> respectively, to line <math>AC</math>. The perimeter of <math>A_1B_1C_1D_1</math> is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
−
==Solution==
+
==Diagram==
 
[[File:Leonard_my_dude's_image.png|center]]
 
[[File:Leonard_my_dude's_image.png|center]]
  
 +
==Solution==
 
Let <math>O</math> be the intersection of <math>AC</math> and <math>BD</math>. Let <math>\theta = \angle AOB</math>.
 
Let <math>O</math> be the intersection of <math>AC</math> and <math>BD</math>. Let <math>\theta = \angle AOB</math>.
  

Revision as of 17:28, 1 April 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

Leonard my dude's image.png

Solution

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

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 $\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.$ 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_11&oldid=150824"