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Difference between revisions of "1978 AHSME Problems/Problem 29"
(→‎Solution)
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<math>\textbf{(A) }20\qquad \textbf{(B) }40\qquad \textbf{(C) }45\qquad \textbf{(D) }50\qquad  \textbf{(E) }60</math>
 
<math>\textbf{(A) }20\qquad \textbf{(B) }40\qquad \textbf{(C) }45\qquad \textbf{(D) }50\qquad  \textbf{(E) }60</math>
 
==Solution==
 
==Solution==
−
It's D, 50
+
 
 +
Diagram:
 +
 
 +
Asymptote code below
 +
[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]
 +
 
 +
Notice that the area of \triangle <math>DAB</math> is the same as that of \triangle <math>A'AB</math> (same base, same height). Thus, the area of \triangle  <math>A'AB</math> is twice that (same height, twice the base). Similarly, [\triangle <math>BB'C</math>] = 2 \cdot [\triangle <math>ABC</math>], and so on.
 +
 
 +
Adding all of these, we see that the area the four triangles around <math>ABCD</math> is twice [\triangle <math>DAB</math>] + [\triangle <math>ABC</math>] + [\triangle <math>BCD</math>] + [\triangle <math>CDA</math>], which is itself twice the area of the quadrilateral <math>ABCD</math>. Finally, [<math>A'B'C'D'</math>] = [<math>ABCD</math>] + 4 \cdot [<math>ABCD</math>] = 5 \cdot [<math>ABCD</math>] = \fbox{50}.
 +
 
 +
~ Mathavi

Revision as of 12:59, 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 $A'$. Also, $AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8$ and $DA = AA' = 9$; and the area of $ABCD$ is $10$. The area of $A 'B 'C'D'$ is

$\textbf{(A) }20\qquad \textbf{(B) }40\qquad \textbf{(C) }45\qquad \textbf{(D) }50\qquad \textbf{(E) }60$

Solution

Diagram:

Asymptote code below [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]

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

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