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Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 9"

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In <math>\triangle</math>ADC, <math>\frac {AE}{ED}</math> = <math>\frac {1}{2}</math> because <math>\frac {[ACE]}{[CED]}</math> = <math>\frac {1}{2}</math> (By Base Division Theorem). <math>[ECD]</math> = <math>2[AEC]</math> (AECF is a parallelogram and <math>2[AEC]</math> = <math>2[AECF]</math>). <math>2[EFC]</math> = <math>[EDC]</math> (same reason as before). <math>[EFC]</math> = <math>\frac {1}{2} * 8 * EC</math>.
+
In <math>\triangle</math>ADC, <math>\frac {AE}{ED}</math> = <math>\frac {1}{2}</math> because <math>\frac {[ACE]}{[CED]}</math> = <math>\frac {1}{2}</math> (By Base Division Theorem). <math>[ECD]</math> = <math>2[AEC]</math> (AECF is a parallelogram and <math>2[AEC]</math> = <math>2[AECF]</math>),
 +
<math>2[EFC]</math> = <math>[EDC]</math> (same reason as before). <math>[EFC]</math> = <math>\frac {1}{2} * 8 * EC</math>.
  
 
== See also ==
 
== See also ==

Revision as of 08:25, 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 $8$ 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]

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 $x^2$ and each portion has area $x^2 \times\frac{1}{3}$ If x is the base of one of the triangles, then the height will be $\frac{2x}{3}$. By the pythaogrean theorem, longer side of the parallelogram has length $\sqrt(x^2+(\frac{2x}{3})^2)$ 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

[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]

In $\triangle$ADC, $\frac {AE}{ED}$ = $\frac {1}{2}$ because $\frac {[ACE]}{[CED]}$ = $\frac {1}{2}$ (By Base Division Theorem). $[ECD]$ = $2[AEC]$ (AECF is a parallelogram and $2[AEC]$ = $2[AECF]$), 

$2[EFC]$ = $[EDC]$ (same reason as before). $[EFC]$ = $\frac {1}{2} * 8 * EC$.

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
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