AoPSWiki

Our Precalculus course starts on Dec. 4. Master trig, complex numbers, and vectors and matrices in 2 and 3 dimensions. Click here to enroll today!

Personal tools

Log in

Search

Toolbox

2007 AIME I Problems/Problem 13

From AoPSWiki

Problem

A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length $4$ . A plane passes through the midpoints of $AE$ , $BC$ , and $CD$ . The plane's intersection with the pyramid has an area that can be expressed as $\sqrt{p}$ . Find $p$ .

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4);triple A=(-2,2,0),...

Solution

Solution 1

Note first that the intersection is a pentagon.

Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\ B(2,2,0),\ C(2,-2,0),\ D(-2,-2,0),\ E(0,0,2\sqrt{2})$ . Using the coordinates of the three points of intersection ( $(-1,1,\sqrt{2}),\ (2,0,0),\ (0,-2,0)$ ), it is possible to determine the equation of the plane. The equation of a plane resembles $ax + by + cz = d$ , and using the points we find that $2a = d \Longrightarrow d = \frac{a}{2}$ , $-2b = d \Longrightarrow d = \frac{-b}{2}$ , and $-a + b + \sqrt{2}c = d \Longrightarrow -\frac{d}{2} - \frac{d}{2} + \sqrt{2}c = d \Longrightarrow c = d\sqrt{2}$ . It is then $x - y + 2\sqrt{2}z = 2$ .

import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4);triple A=(-2,2,0),...

pointpen = black; pathpen = black+linewidth(0.7); pair P = (0, 2.5^.5), X = (3/2^.5,0), Y = (-3/2^.5,0), Q = (2^.5,-2.5^.5), ...

Write the equation of the lines and substitute to find that the other two points of intersection on $\overline{BE}$ , $\overline{DE}$ are $\left(\frac{\pm 3}{2},\frac{\pm 3}{2},\frac{\sqrt{2}}{2}\right)$ . To find the area of the pentagon, break it up into pieces (an isosceles triangle on the top, an isosceles trapezoid on the bottom). Using the distance formula ( $\sqrt{a^2 + b^2 + c^2}$ ), it is possible to find that the area of the triangle is $\frac{1}{2}bh \Longrightarrow \frac{1}{2} 3\sqrt{2} \cdot \sqrt{\frac 52} = \frac{3\sqrt{5}}{2}$ . The trapezoid has area $\frac{1}{2}h(b_1 + b_2) \Longrightarrow \frac 12\sqrt{\frac 52}\left(2\sqrt{2} + 3\sqrt{2}\right) = \frac{5\sqrt{5}}{2}$ . In total, the area is $4\sqrt{5} = \sqrt{80}$ , and the solution is $\boxed{080}$ .

Solution 2

Use the same coordinate system as above, and let the plane determined by $\triangle PQR$ intersect $\overline{BE}$ at $X$ and $\overline{DE}$ at $Y$ . Then the line $\overline{XY}$ is the intersection of the planes determined by $\triangle PQR$ and $\triangle BDE$ .

Note that the plane determined by $\triangle BDE$ has the equation $x=y$ , and $\overline{PQ}$ can be described by $x=2(1-t)-t,\ y=t,\ z=t\sqrt{2}$ . It intersects the plane when $2(1-t)-t=t$ , or $t=\frac{1}{2}$ . This intersection point has $z=\frac{\sqrt{2}}{2}$ . Similarly, the intersection between $\overline{PR}$ and $\triangle BDE$ has $z=\frac{\sqrt{2}}{2}$ . So $\overline{XY}$ lies on the plane $z=\frac{\sqrt{2}}{2}$ , from which we obtain $X=\left( \frac{3}{2},\frac{3}{2},\frac{\sqrt{2}}{2}\right)$ and $Y=\left( -\frac{3}{2},-\frac{3}{2},\frac{\sqrt{2}}{2}\right)$ . The area of the pentagon $EXQRY$ can be computed in the same way as above.

See also

2007 AIME I (Problems • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems/Problem_13"

Category: Intermediate Geometry Problems

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us