Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 9"
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Thus, the area of the square is 64*13 = 832. | Thus, the area of the square is 64*13 = 832. | ||
| − | == Solution 2 ( | + | == Solution 2 (ArGKK) == |
<asy> | <asy> | ||
unitsize(135); | unitsize(135); | ||
Revision as of 07:56, 28 September 2023
Problem
A square is divided into three pieces of
equal area by two parallel lines as shown.
If the distance between the two parallel
lines is 
what is the area of the square?
![[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); draw((1,0)--(0,2/3),black); draw((1,1/3)--(0,1),black); [/asy]](http://latex.artofproblemsolving.com/5/9/4/594509287728bec7666a5a011722c3330803fba0.png)
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Let x be the length of a side. Then the square has area 
and each portion has area 
If x is the base of one of the triangles, then the height will be 
.
By the pythaogrean theorem, longer side of the parallelogram has length 
Thus sqrt(13)*x/3*8 = x^2/3. Solving this gives x = 8*sqrt(13).
Thus, the area of the square is 64*13 = 832.
Solution 2 (ArGKK)
![[asy] unitsize(135); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A, B, C, D, E, F; A=(0, 0); B=(1, 0); C=(1, -1); D=(0,-1); E=(0,-1/3); F=(1,-2/3); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$E$",E,W); label("$F$",F,SE); draw(A--F); draw(C--E); draw(A--C); draw(E--F); [/asy]](http://latex.artofproblemsolving.com/7/3/3/7332665745d0c13c92b6188693abd3e776e1ff4b.png)
In 
ADC, 
= 
because ![$\frac {[ACE]}{[CED]}$](http://latex.artofproblemsolving.com/6/b/7/6b78791905d0e5e4051669f9c5e57f65aed58f51.png)
= (According to question)
![$[ECD]$](http://latex.artofproblemsolving.com/0/8/7/087b2e05f78df428cf0babd9f2a63bf8a00e16b3.png)
= ![$2[EFC]$](http://latex.artofproblemsolving.com/d/0/7/d07106120932893c0945f6e4b7c5517e297d9f91.png)
(AECF is a parallelogram and = ![$[AECF]$](http://latex.artofproblemsolving.com/3/0/5/30513ebafd266f02c11199ee20ec95d7c14260d9.png)
)
= ![$[EDC]$](http://latex.artofproblemsolving.com/b/d/a/bda1c8e25d3765b68beb4f80e545d40538fc376d.png)
(same reason as before)
![$[EFC]$](http://latex.artofproblemsolving.com/3/c/2/3c260ecefa4495de05e121a3c5addeda4ebe2950.png)
= * 8 * EC
= 
* EC * (Perpendicular from D to EC)
2 * * 8 * EC = * EC * (Perpendicular from D to EC)
(Perpendicular from D to EC) = 16
Let AE = 
, then ED = 
. So the side length of square is 
.
(Perpendicular from D to EC) = 
EC = 
(By Pythagoras Theorem)
16 = 
16 = 
= 
= 
Area of square ABCD = 
= 
= 832
See also
| 2009 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
