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 "2000 AIME I Problems/Problem 2"

(→‎Solution 1)
(→‎Solution 2)
 
(3 intermediate revisions by 2 users not shown)
Line 2: Line 2:
 
Let <math>u</math> and <math>v</math> be [[integer]]s satisfying <math>0 < v < u</math>. Let <math>A = (u,v)</math>, let <math>B</math> be the [[reflection]] of <math>A</math> across the line <math>y = x</math>, let <math>C</math> be the reflection of <math>B</math> across the y-axis, let <math>D</math> be the reflection of <math>C</math> across the x-axis, and let <math>E</math> be the reflection of <math>D</math> across the y-axis. The area of [[pentagon]] <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>.
 
Let <math>u</math> and <math>v</math> be [[integer]]s satisfying <math>0 < v < u</math>. Let <math>A = (u,v)</math>, let <math>B</math> be the [[reflection]] of <math>A</math> across the line <math>y = x</math>, let <math>C</math> be the reflection of <math>B</math> across the y-axis, let <math>D</math> be the reflection of <math>C</math> across the x-axis, and let <math>E</math> be the reflection of <math>D</math> across the y-axis. The area of [[pentagon]] <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>.
  
−
== Solution ==
+
== Solutions ==
 
=== Solution 1 ===
 
=== Solution 1 ===
 
<center><asy>
 
<center><asy>
Line 14: Line 14:
  
 
=== Solution 2 ===
 
=== Solution 2 ===
 +
We find the coordinates like in the solution above: <math>A = (u,v)</math>, <math>B = (v,u)</math>, <math>C = (-v,u)</math>, <math>D = (-v,-u)</math>, <math>E = (v,-u)</math>. Then we notice pentagon <math>ABCDE</math> fits into a rectangle of side lengths <math>(u+v)</math> and <math>(2u)</math>, giving us two triangles, each with hypotenuse <math>AB</math> and <math>BE</math>. First, we can solve for the first triangle. Using the coordinates of <math>A</math> and <math>B</math>, we discover the side lengths are both <math>(u-v)</math>, so the area of the triangle of hypotenuse <math>AB</math> is <math>\frac{1}{2}(u-v)^2</math>. Next, we can solve for the second triangle. Using the coordinates of <math>A</math> and <math>E</math>, we discover the side lengths are <math>(u-v)</math> and <math>(u+v)</math>, so the area of the triangle of hypotenuse <math>AE</math> is <math>\frac{1}{2}(u-v)(u+v) = \frac{1}{2}(u^2-v^2)</math>. Now, let’s subtract the area of these 2 triangles from the rectangle giving us <math>(u+v)(2u)-\frac{1}{2}(u-v)^2-\frac{1}{2}(u^2-v^2)=451 —> u^2+3uv=451 -> u(u+3v)=451</math>. Next, we take note of the fact that <math>u</math> and <math>u+3v</math> are both factors of 451, and since both <math>u</math> and <math>v</math> are positive integers, <math>u+3v</math> must be greater than <math>u</math>, thus giving us two cases, where either <math>u=1</math> or <math>u=11</math>. After trying both, the only working pair of <math>(u,v)</math> where both <math>u</math> and <math>v</math> are integers are <math>u=11</math> and <math>v=10</math>, thus meaning <math>u + v =</math> <math>\boxed{021}</math>
 +
 +
~Aeioujyot
 +
 +
=== Solution 3 ===
 
We find the coordinates like in the solution above: <math>A = (u,v)</math>, <math>B = (v,u)</math>, <math>C = (-v,u)</math>, <math>D = (-v,-u)</math>, <math>E = (v,-u)</math>. Then we apply the [[Shoelace Theorem]]. <cmath>A =  \frac{1}{2}[(u^2 + vu + vu + vu + v^2) - (v^2 - uv - uv - uv -u^2)] = 451</cmath> <cmath>\frac{1}{2}(2u^2 + 6uv) = 451</cmath> <cmath>u(u + 3v) = 451</cmath>
 
We find the coordinates like in the solution above: <math>A = (u,v)</math>, <math>B = (v,u)</math>, <math>C = (-v,u)</math>, <math>D = (-v,-u)</math>, <math>E = (v,-u)</math>. Then we apply the [[Shoelace Theorem]]. <cmath>A =  \frac{1}{2}[(u^2 + vu + vu + vu + v^2) - (v^2 - uv - uv - uv -u^2)] = 451</cmath> <cmath>\frac{1}{2}(2u^2 + 6uv) = 451</cmath> <cmath>u(u + 3v) = 451</cmath>
  

Latest revision as of 13:31, 22 October 2024

Problem

Let $u$ and $v$ be integers satisfying $0 < v < u$. Let $A = (u,v)$, let $B$ be the reflection of $A$ across the line $y = x$, let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$. Find $u + v$.

Solutions

Solution 1

[asy] pointpen = black; pathpen = linewidth(0.7) + black; size(180); pair A=(11,10), B=(10,11), C=(-10, 11), D=(-10, -11), E=(10, -11); D(D(MP("A\ (u,v)",A,(1,0)))--D(MP("B",B,N))--D(MP("C",C,N))--D(MP("D",D))--D(MP("E",E))--cycle); D((-15,0)--(15,0),linewidth(0.6),Arrows(5)); D((0,-15)--(0,15),linewidth(0.6),Arrows(5)); D((-15,-15)--(15,15),linewidth(0.6),Arrows(5)); [/asy]

Since $A = (u,v)$, we can find the coordinates of the other points: $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)(2v) = 4uv$ and the area of $ABE$ is $\frac{1}{2}(2u)(u-v) = u^2 - uv$. Adding these together, we get $u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41$. Since $u,v$ are positive, $u+3v>u$, and by matching factors we get either $(u,v) = (1,150)$ or $(11,10)$. Since $v < u$ the latter case is the answer, and $u+v = \boxed{021}$.

Solution 2

We find the coordinates like in the solution above: $A = (u,v)$, $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. Then we notice pentagon $ABCDE$ fits into a rectangle of side lengths $(u+v)$ and $(2u)$, giving us two triangles, each with hypotenuse $AB$ and $BE$. First, we can solve for the first triangle. Using the coordinates of $A$ and $B$, we discover the side lengths are both $(u-v)$, so the area of the triangle of hypotenuse $AB$ is $\frac{1}{2}(u-v)^2$. Next, we can solve for the second triangle. Using the coordinates of $A$ and $E$, we discover the side lengths are $(u-v)$ and $(u+v)$, so the area of the triangle of hypotenuse $AE$ is $\frac{1}{2}(u-v)(u+v) = \frac{1}{2}(u^2-v^2)$. Now, let’s subtract the area of these 2 triangles from the rectangle giving us $(u+v)(2u)-\frac{1}{2}(u-v)^2-\frac{1}{2}(u^2-v^2)=451 —> u^2+3uv=451 -> u(u+3v)=451$. Next, we take note of the fact that $u$ and $u+3v$ are both factors of 451, and since both $u$ and $v$ are positive integers, $u+3v$ must be greater than $u$, thus giving us two cases, where either $u=1$ or $u=11$. After trying both, the only working pair of $(u,v)$ where both $u$ and $v$ are integers are $u=11$ and $v=10$, thus meaning $u + v =$ $\boxed{021}$

~Aeioujyot

Solution 3

We find the coordinates like in the solution above: $A = (u,v)$, $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. Then we apply the Shoelace Theorem. \[A = \frac{1}{2}[(u^2 + vu + vu + vu + v^2) - (v^2 - uv - uv - uv -u^2)] = 451\] \[\frac{1}{2}(2u^2 + 6uv) = 451\] \[u(u + 3v) = 451\]

This means that $(u,v) = (11, 10)$ or $(1,150)$, but since $v < u$, then the answer is $\boxed{021}$

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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=2000_AIME_I_Problems/Problem_2&oldid=230348"