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Difference between revisions of "1978 AHSME Problems/Problem 29"

(Undo revision 177406 by Mathavi (talk))
(Tag: Undo)
(Undo revision 177406 by Mathavi (talk))
(Tag: Undo)
Line 3: Line 3:
  
 
<math></math>
 
<math></math>
 +
[asy]
 +
unitsize(1 cm);
 +
 +
pair[] A, B, C, D;
 +
 +
A[0] = (0,0);
 +
B[0] = (0.6,1.2);
 +
C[0] = (-0.3,2.5);
 +
D[0] = (-1.5,0.7);
 +
B[1] = interp(A[0],B[0],2);
 +
C[1] = interp(B[0],C[0],2);
 +
D[1] = interp(C[0],D[0],2);
 +
A[1] = interp(D[0],A[0],2);
 +
 +
draw(A[1]--B[1]--C[1]--D[1]--cycle);
 +
draw(A[0]--B[1]);
 +
draw(B[0]--C[1]);
 +
draw(C[0]--D[1]);
 +
draw(D[0]--A[1]);
 +
 +
label("<math>A</math>", A[0], SW);
 +
label("<math>B</math>", B[0], SE);
 +
label("<math>C</math>", C[0], NE);
 +
label("<math>D</math>", D[0], NW);
 +
label("<math>A'</math>", A[1], SE);
 +
label("<math>B'</math>", B[1], NE);
 +
label("<math>C'</math>", C[1], N);
 +
label("<math>D'</math>", D[1], SW);
 +
 +
[\asy]
 +
<cmath>
 +
 +
</cmath>
 
[asy]
 
[asy]
 
unitsize(1 cm);
 
unitsize(1 cm);

Revision as of 13:20, 23 August 2022

Problem

Sides $AB$, $BC$, $CD$, and $DA$, respectively of convex quadrilateral $ABCD$ are extended past $B$, $C$, $D$, and $A$ to points $B'$, $C'$, $D'$, and $A'$. If $AB = BB' = 6$, $BC = CC' = 7$, $CD = DD' = 8$, and $DA = AA' = 9$, and the area of $ABCD$ is 10, determine the area of quadrilateral $A'B'C'D'$.

$$ (Error compiling LaTeX. Unknown error_msg) [asy] unitsize(1 cm);

pair[] A, B, C, D;

A[0] = (0,0); B[0] = (0.6,1.2); C[0] = (-0.3,2.5); D[0] = (-1.5,0.7); B[1] = interp(A[0],B[0],2); C[1] = interp(B[0],C[0],2); D[1] = interp(C[0],D[0],2); A[1] = interp(D[0],A[0],2);

draw(A[1]--B[1]--C[1]--D[1]--cycle); draw(A[0]--B[1]); draw(B[0]--C[1]); draw(C[0]--D[1]); draw(D[0]--A[1]);

label("$A$", A[0], SW); label("$B$", B[0], SE); label("$C$", C[0], NE); label("$D$", D[0], NW); label("$A'$", A[1], SE); label("$B'$", B[1], NE); label("$C'$", C[1], N); label("$D'$", D[1], SW);

[\asy] \[\] [asy] unitsize(1 cm);

pair[] A, B, C, D;

A[0] = (0,0); B[0] = (0.6,1.2); C[0] = (-0.3,2.5); D[0] = (-1.5,0.7); B[1] = interp(A[0],B[0],2); C[1] = interp(B[0],C[0],2); D[1] = interp(C[0],D[0],2); A[1] = interp(D[0],A[0],2);

draw(A[1]--B[1]--C[1]--D[1]--cycle); draw(A[0]--B[1]); draw(B[0]--C[1]); draw(C[0]--D[1]); draw(D[0]--A[1]);

label("$A$", A[0], SW); label("$B$", B[0], SE); label("$C$", C[0], NE); label("$D$", D[0], NW); label("$A'$", A[1], SE); label("$B'$", B[1], NE); label("$C'$", C[1], N); label("$D'$", D[1], SW);

[\asy] $$ (Error compiling LaTeX. Unknown error_msg)

Solution

Notice that the area of $\triangle$ $DAB$ is the same as that of $\triangle$ $A'AB$ (same base, same height). Thus, the area of $\triangle$ $A'AB$ is twice that (same height, twice the base). Similarly, [$\triangle$ $BB'C$] = 2 $\cdot$ [$\triangle$ $ABC$], and so on.

Adding all of these, we see that the area the four triangles around $ABCD$ is twice [$\triangle$ $DAB$] + [$\triangle$ $ABC$] + [$\triangle$ $BCD$] + [$\triangle$ $CDA$], which is itself twice the area of the quadrilateral $ABCD$. Finally, [$A'B'C'D'$] = [$ABCD$] + 4 $\cdot$ [$ABCD$] = 5 $\cdot$ [$ABCD$] = $\fbox{50}$.

~ Mathavi

Note: Anyone with a diagram would be of great help (still new to LaTex).

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