Difference between revisions of "2001 AIME I Problems/Problem 13"

Revision as of 05:07, 18 June 2020

Problem

In a certain circle, the chord of a $d$ -degree arc is $22$ centimeters long, and the chord of a $2d$ -degree arc is $20$ centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$

Solution

Solution 1

Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$ -degree arcs and one chord of one $3d$ -degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$ -degree arcs. Let $AB$ , $AC$ , and $BD$ be the chords of the $d$ -degree arcs, and let $CD$ be the chord of the $3d$ -degree arc. Also let $x$ be equal to the chord length of the $3d$ -degree arc. Hence, the length of the chords, $AD$ and $BC$ , of the $2d$ -degree arcs can be represented as $x + 20$ , as given in the problem.

Using Ptolemy's theorem,

$AB(CD) + AC(BD) = AD(BC)$ $22x + 22(22) = (x + 20)^2$ $22x + 484 = x^2 + 40x + 400$ $0 = x^2 + 18x - 84$

We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. $x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}$ $x = \frac{-18 + \sqrt{660}}{2}$

$x$ simplifies to $\frac{-18 + 2\sqrt{165}}{2},$ which equals $-9 + \sqrt{165}.$ Thus, the answer is $9 + 165 = \boxed{174}$ .

Solution 2

Let $z=\frac{d}{2},$ and $R$ be the circumradius. From the given information, $2R\sin z=22$ $2R(\sin 2z-\sin 3z)=20$ Dividing the latter by the former, $\frac{2\sin z\cos z-(3\cos^2z\sin z-\sin^3 z)}{\sin z}=2\cos z-(3\cos^2z-\sin^2z)=1+2\cos z-4\cos^2z=\frac{10}{11}$ $4\cos^2z-2\cos z-\frac{1}{11}=0 (1)$ We want to find $\frac{22\sin (3z)}{\sin z}=22(3-4\sin^2z)=22(4\cos^2z-1).$ From $(1),$ this is equivalent to $44\cos z-20.$ Using the quadratic formula, we find that the desired length is equal to $\sqrt{165}-9,$ so our answer is $\boxed{174.}$

Solution 3

Let $z=\frac{d}{2}$ , R $be the circumradius, and$ a $be the length of 3d degree chord. Using the extended sine law, we obtain: <cmath>22=2Rsin(z)</cmath> <cmath>20+a=2Rsin(2z)</cmath> <cmath>a=2Rsin(3z)</cmath> Dividing the second from the first we get$ cos(z)=\frac{20+a}{44} $By the triple angle formula we can manipulate the third equation as follows:$ a=2R\times sin(3z)=\frac{22}{sin(z)} \times (3sin(z)-4sin^3(z)) = 22(3-4sin^2(z))=22(4cos^2(z))=\frac{(20+a)^2}{22}-22$ Solving the quadratic equation gives the answer.

@@ Line 28: / Line 28: @@
 ===Solution 3===
-Let <math>z=\frac{d}{2},</math> R<math> be the circumradius, and </math>a<math> be the length of 3d degree chord. Using the extended sine law, we obtain:
+Let <math>z=\frac{d}{2}</math>, R<math> be the circumradius, and </math>a<math> be the length of 3d degree chord. Using the extended sine law, we obtain:
 <cmath>22=2Rsin(z)</cmath>
 <cmath>20+a=2Rsin(2z)</cmath>
 <cmath>a=2Rsin(3z)</cmath>
 Dividing the second from the first we get </math>cos(z)=\frac{20+a}{44}<math>
-By the triple angle formula we can manipulate the third equation as follows:
+By the triple angle formula we can manipulate the third equation as follows: </math>a=2R\times sin(3z)=\frac{22}{sin(z)} \times (3sin(z)-4sin^3(z)) = 22(3-4sin^2(z))=22(4cos^2(z))=\frac{(20+a)^2}{22}-22$
-</math>a=2R\times sin(3z)=\frac{22}{sin(z)} \times (3sin(z)-4sin^3(z)) = 22(3-4sin^2(z))=22(4cos^2(z))$
+Solving the quadratic equation gives the answer.
-Solving it gives the answer.
 == See also ==

2001 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search