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Difference between revisions of "1989 AHSME Problems/Problem 21"
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<asy>
 
<asy>
−
draw((0,0)--(0,5)--(5,5)--(5,0)--(0,0));
+
unitsize(2.5 cm);
−
draw((0,1)--(4,5));
+
 
−
draw((1,0)--(5,4));
+
pair[] A, B, C;
−
draw((0,4)--(4,0));
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real t = 0.2;
−
draw((1,5)--(5,1));
+
 
−
label("blue",(2.5,2.5));
+
A[1] = (0,0);
−
label("red",(1,1));
+
A[2] = (1,0);
−
label("red",(1,4));
+
A[3] = (1,1);
−
label("red",(4,1));
+
A[4] = (0,1);
−
label("red",(4,4));
+
B[1] = (t,0);
 +
B[2] = (1 - t,0);
 +
B[3] = (1,t);
 +
B[4] = (1,1 - t);
 +
B[5] = (1 - t,1);
 +
B[6] = (t,1);
 +
B[7] = (0,1 - t);
 +
B[8] = (0,t);
 +
C[1] = extension(B[1],B[4],B[7],B[2]);
 +
C[2] = extension(B[3],B[6],B[1],B[4]);
 +
C[3] = extension(B[5],B[8],B[3],B[6]);
 +
C[4] = extension(B[7],B[2],B[5],B[8]);
 +
 
 +
fill(C[1]--C[2]--C[3]--C[4]--cycle,blue);
 +
fill(A[1]--B[1]--C[1]--C[4]--B[8]--cycle,red);
 +
fill(A[2]--B[3]--C[2]--C[1]--B[2]--cycle,red);
 +
fill(A[3]--B[5]--C[3]--C[2]--B[4]--cycle,red);
 +
fill(A[4]--B[7]--C[4]--C[3]--B[6]--cycle,red);
 +
 
 +
draw(A[1]--A[2]--A[3]--A[4]--cycle);
 +
draw(B[1]--B[4]);
 +
draw(B[2]--B[7]);
 +
draw(B[3]--B[6]);
 +
draw(B[5]--B[8]);
 
</asy>
 
</asy>
  
−
<math>\text{(A)}}\ .5\qquad\text{(B)}\ 1\qquad\text{(C)}\ 2\qquad\text{(D)}\ 3\qquad\text{(E)}\ 6</math>
+
<math>\text{(A)}\ 0.5\qquad\text{(B)}\ 1\qquad\text{(C)}\ 2\qquad\text{(D)}\ 3\qquad\text{(E)}\ 6</math>
 +
 
 
==Solution==
 
==Solution==
 
The diagram can be quartered as shown:
 
The diagram can be quartered as shown:
Line 39: Line 63:
 
</asy>
 
</asy>
 
The border in this figure is the former cross, which still occupies 36% of the area. Therefore the inner square occupies 64% of the area, from which we deduce that it is <math>0.8k \times 0.8k</math>, and that one blue square must be <math>0.1k\times 0.1k=0.01k^2</math> or 1% each. Thus the blue area is <math>\boxed{2\%}</math> of the total.
 
The border in this figure is the former cross, which still occupies 36% of the area. Therefore the inner square occupies 64% of the area, from which we deduce that it is <math>0.8k \times 0.8k</math>, and that one blue square must be <math>0.1k\times 0.1k=0.01k^2</math> or 1% each. Thus the blue area is <math>\boxed{2\%}</math> of the total.
 +
 +
 +
== See also ==
 +
{{AHSME box|year=1989|num-b=20|num-a=22}} 
 +
 +
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:59, 1 April 2020

Problem

A square flag has a red cross of uniform width with a blue square in the center on a white background as shown. (The cross is symmetric with respect to each of the diagonals of the square.) If the entire cross (both the red arms and the blue center) takes up 36% of the area of the flag, what percent of the area of the flag is blue?

[asy] unitsize(2.5 cm); pair[] A, B, C; real t = 0.2; A[1] = (0,0); A[2] = (1,0); A[3] = (1,1); A[4] = (0,1); B[1] = (t,0); B[2] = (1 - t,0); B[3] = (1,t); B[4] = (1,1 - t); B[5] = (1 - t,1); B[6] = (t,1); B[7] = (0,1 - t); B[8] = (0,t); C[1] = extension(B[1],B[4],B[7],B[2]); C[2] = extension(B[3],B[6],B[1],B[4]); C[3] = extension(B[5],B[8],B[3],B[6]); C[4] = extension(B[7],B[2],B[5],B[8]); fill(C[1]--C[2]--C[3]--C[4]--cycle,blue); fill(A[1]--B[1]--C[1]--C[4]--B[8]--cycle,red); fill(A[2]--B[3]--C[2]--C[1]--B[2]--cycle,red); fill(A[3]--B[5]--C[3]--C[2]--B[4]--cycle,red); fill(A[4]--B[7]--C[4]--C[3]--B[6]--cycle,red); draw(A[1]--A[2]--A[3]--A[4]--cycle); draw(B[1]--B[4]); draw(B[2]--B[7]); draw(B[3]--B[6]); draw(B[5]--B[8]); [/asy]

$\text{(A)}\ 0.5\qquad\text{(B)}\ 1\qquad\text{(C)}\ 2\qquad\text{(D)}\ 3\qquad\text{(E)}\ 6$

Solution

The diagram can be quartered as shown: [asy] draw((0,0)--(0,5)--(5,5)--(5,0)--(0,0)); draw((0,1)--(4,5)); draw((1,0)--(5,4)); draw((0,4)--(4,0)); draw((1,5)--(5,1)); draw((0,0)--(5,5),dotted); draw((0,5)--(5,0),dotted); [/asy] and reassembled into two smaller squares of side $k$, each of which looks like this: [asy] draw((0,0)--(0,5)--(5,5)--(5,0)--(0,0)); draw((0,1)--(4,1)--(4,5)); draw((1,0)--(1,4)--(5,4)); label("blue",(0.5,0.5)); label("blue",(4.5,4.5)); label("red",(0.5,4.5)); label("red",(4.5,0.5)); label("white",(2.5,2.5)); [/asy] The border in this figure is the former cross, which still occupies 36% of the area. Therefore the inner square occupies 64% of the area, from which we deduce that it is $0.8k \times 0.8k$, and that one blue square must be $0.1k\times 0.1k=0.01k^2$ or 1% each. Thus the blue area is $\boxed{2\%}$ of the total.


See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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