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 "1998 AIME Problems/Problem 11"

m (→‎Solution 1)
m (→‎Solution 2)
Line 59: Line 59:
 
Now, the next vertex (call it <math>Y</math>) will be somewhere on <math>AF</math>.  The segment must be parallel to <math>QR</math>, so <math>FY</math> must have length <math>2x</math>,  and <math>AY</math> must be <math>20 - 2x</math>.
 
Now, the next vertex (call it <math>Y</math>) will be somewhere on <math>AF</math>.  The segment must be parallel to <math>QR</math>, so <math>FY</math> must have length <math>2x</math>,  and <math>AY</math> must be <math>20 - 2x</math>.
  
−
Since <math>DX \parallel AP</math>, <math>DR \parallel AY</math>, and <math>RX \parallel PY</math>, we must have <math>\triangle APY\sim\triangleDXR</math>; therefore,
+
Since <math>DX \parallel AP</math>, <math>DR \parallel AY</math>, and <math>RX \parallel PY</math>, we must have <math>\triangle{APY} \sim \triangle{DXR}</math>; therefore, <cmath>\frac{AP}{DX}=\frac{AY}{DR}</cmath> <cmath>\frac{5}{x}=\frac{20-2x}{10}</cmath> <cmath>x^{2}-10x+25=0</cmath> <cmath>x=5</cmath>
−
 
 
−
<div style="text-align:center;">
 
−
<math>\frac{AP}{DX}=\frac{AY}{DR}</math><br />
 
−
<math>\frac{5}{x}=\frac{20-2x}{10}</math><br />
 
−
<math>x^{2}-10x+25=0</math><br />
 
−
<math>x=5</math></div>
 
  
 
We can now find that the hexagon has side lengths <math>15\sqrt {2}</math>, <math>5\sqrt {5}</math>, <math>5\sqrt {5}</math>, <math>15\sqrt {2}</math>, <math>5\sqrt {5}</math>, and <math>5\sqrt {5}</math>.  Moreover, opposite angles of this must be equal (by symmetry), so segment <math>RY</math> divides the [[hexagon]] into two [[isosceles trapezoid]]s.  It is easy to find the length of <math>RY</math> (they're midpoints of opposite edges, so the distance between the two points is equal to a face diagonal of the cube, or <math>20\sqrt {2}</math>), so it is now easy to finish the problem. From here, we can continue as in the first solution.
 
We can now find that the hexagon has side lengths <math>15\sqrt {2}</math>, <math>5\sqrt {5}</math>, <math>5\sqrt {5}</math>, <math>15\sqrt {2}</math>, <math>5\sqrt {5}</math>, and <math>5\sqrt {5}</math>.  Moreover, opposite angles of this must be equal (by symmetry), so segment <math>RY</math> divides the [[hexagon]] into two [[isosceles trapezoid]]s.  It is easy to find the length of <math>RY</math> (they're midpoints of opposite edges, so the distance between the two points is equal to a face diagonal of the cube, or <math>20\sqrt {2}</math>), so it is now easy to finish the problem. From here, we can continue as in the first solution.

Revision as of 19:37, 11 March 2015

Problem

Three of the edges of a cube are $\overline{AB}, \overline{BC},$ and $\overline{CD},$ and $\overline{AD}$ is an interior diagonal. Points $P, Q,$ and $R$ are on $\overline{AB}, \overline{BC},$ and $\overline{CD},$ respectively, so that $AP = 5, PB = 15, BQ = 15,$ and $CR = 10.$ What is the area of the polygon that is the intersection of plane $PQR$ and the cube?

Solution

Solution 1

For non-asymptote version of image, see Image:1998_AIME-11.png

[asy] import three; size(280); defaultpen(linewidth(0.6)+fontsize(9)); currentprojection=perspective(30,-60,40); triple A=(0,0,0),B=(20,0,0),C=(20,0,20),D=(20,20,20); triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0); draw(box((0,0,0),(20,20,20))); draw(P--Q--R--Pa--Qa--Ra--cycle,linewidth(0.7)); label("\(A\,(0,0,0)\)",A,SW); label("\(B\,(20,0,0)\)",B,S); label("\(C\,(20,0,20)\)",C,SW); label("\(D\,(20,20,20)\)",D,E); label("\(P\,(5,0,0)\)",P,SW); label("\(Q\,(20,0,15)\)",Q,E); label("\(R\,(20,10,20)\)",R,E); label("\((15,20,20)\)",Pa,N); label("\((0,20,5)\)",Qa,W); label("\((0,10,0)\)",Ra,W); [/asy]

This approach uses analytical geometry. Let $A$ be at the origin, $B$ at $(20,0,0)$, $C$ at $(20,0,20)$, and $D$ at $(20,20,20)$. Thus, $P$ is at $(5,0,0)$, $Q$ is at $(20,0,15)$, and $R$ is at $(20,10,20)$.

Let the plane $PQR$ have the equation $ax + by + cz = d$. Using point $P$, we get that $5a = d$. Using point $Q$, we get $20a + 15c = d \Longrightarrow 4d + 15c = d \Longrightarrow d = -5c$. Using point $R$, we get $20a + 10b + 20c = d \Longrightarrow 4d + 10b - 4d = d \Longrightarrow d = 10b$. Thus plane $PQR$’s equation reduces to $\frac{d}{5}x + \frac{d}{10}y - \frac{d}{5}z = d \Longrightarrow 2x + y - 2z = 10$.

We know need to find the intersection of this plane with that of $z = 0$, $z = 20$, $x = 0$, and $y = 20$. After doing a little bit of algebra, the intersections are the lines $y = -2x + 10$, $y = -2x + 50$, $y = 2z + 10$, and $z = x + 5$. Thus, there are three more vertices on the polygon, which are at $(0,10,0)(0,20,5)(15,20,20)$.

We can find the lengths of the sides of the polygons now. There are 4 right triangles with legs of length 5 and 10, so their hypotenuses are $5\sqrt{5}$. The other two are of $45-45-90 \triangle$s with legs of length 15, so their hypotenuses are $15\sqrt{2}$. So we have a hexagon with sides $15\sqrt{2},5\sqrt{5}, 5\sqrt{5},15\sqrt{2}, 5\sqrt{5},5\sqrt{5}$ By symmetry, we know that opposite angles of the polygon are congruent. We can also calculate the length of the long diagonal by noting that it is of the same length of a face diagonal, making it $20\sqrt{2}$.

[asy] size(190); pointpen=black;pathpen=black; real s=2^.5; pair P=(0,0),Q=(7.5*s,2.5*s),R=Q+(0,15*s),Pa=(0,20*s),Qa=(-Q.x,Q.y),Ra=(-R.x,R.y); D(P--Q--R--Pa--Ra--Qa--cycle);D(R--Ra);D(Q--Qa);D(P--Pa); MP("15\sqrt{2}",(Q+R)/2,E); MP("5\sqrt{5}",(P+Q)/2,SE); MP("5\sqrt{5}",(R+Pa)/2,NE); MP("20\sqrt{2}",(P+Pa)/2,W); [/asy]

The height of the triangles at the top/bottom is $\frac{20\sqrt{2} - 15\sqrt{2}}{2} = \frac{5}{2}\sqrt{2}$. The Pythagorean Theorem gives that half of the base of the triangles is $\frac{15}{\sqrt{2}}$. We find that the middle rectangle is actually a square, so the total area is $(15\sqrt{2})^2 + 4\left(\frac 12\right)\left(\frac 52\sqrt{2}\right)\left(\frac{15}{\sqrt{2}}\right) = 525$.

Solution 2

First, note that whenever the plane intersects two opposite faces of the cube, the resulting line segments must be parallel. Because they are part of parallel planes (the faces), they must be either parallel or skew; they are both part of plane $PQR$, so they cannot be skew. Therefore, they are parallel.

Let the cube's vertices be $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$, with $A$, $B$, and $C$ on the bottom face as before, $H$ being the other bottom vertex, $D$ directly above $C$, $E$ above $B$, $F$ above $A$, and $G$ above $H$.

Clearly, the next vertex of the intersection (starting with $P$, $Q$, $R$) will be somewhere on $DG$. Let it be $X$, and have a distance of $x$ from D, and a distance of $20 - x$ from $G$.

Then, the next vertex will be somewhere on $FG$. It must be parallel to $PQ$, so this implies that it has a distance of $20 - x$ from $G$, and thus a distance of $x$ from $F$.

Now, the next vertex (call it $Y$) will be somewhere on $AF$. The segment must be parallel to $QR$, so $FY$ must have length $2x$, and $AY$ must be $20 - 2x$.

Since $DX \parallel AP$, $DR \parallel AY$, and $RX \parallel PY$, we must have $\triangle{APY} \sim \triangle{DXR}$; therefore, \[\frac{AP}{DX}=\frac{AY}{DR}\] \[\frac{5}{x}=\frac{20-2x}{10}\] \[x^{2}-10x+25=0\] \[x=5\]

We can now find that the hexagon has side lengths $15\sqrt {2}$, $5\sqrt {5}$, $5\sqrt {5}$, $15\sqrt {2}$, $5\sqrt {5}$, and $5\sqrt {5}$. Moreover, opposite angles of this must be equal (by symmetry), so segment $RY$ divides the hexagon into two isosceles trapezoids. It is easy to find the length of $RY$ (they're midpoints of opposite edges, so the distance between the two points is equal to a face diagonal of the cube, or $20\sqrt {2}$), so it is now easy to finish the problem. From here, we can continue as in the first solution.

See also

1998 AIME (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=1998_AIME_Problems/Problem_11&oldid=68755"