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

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Sides <math>AB,~ BC, ~CD</math> and <math>DA</math>, respectively, of convex quadrilateral <math>ABCD</math> are extended past <math>B,~ C ,~ D</math> and <math>A</math> to points <math>B',~C',~ D'</math> and <math>A'</math>. Also, <math>AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8</math> and <math>DA = AA' = 9</math>; and the area of <math>ABCD</math> is <math>10</math>. The area of <math>A 'B 'C'D'</math> is
 
Sides <math>AB,~ BC, ~CD</math> and <math>DA</math>, respectively, of convex quadrilateral <math>ABCD</math> are extended past <math>B,~ C ,~ D</math> and <math>A</math> to points <math>B',~C',~ D'</math> and <math>A'</math>. Also, <math>AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8</math> and <math>DA = AA' = 9</math>; and the area of <math>ABCD</math> is <math>10</math>. The area of <math>A 'B 'C'D'</math> is
  
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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>
 
 
==Solution==
 
==Solution==
  
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Diagram:
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Notice that the area of <math>\triangle</math> <math>DAB</math> is the same as that of <math>\triangle</math> <math>A'AB</math> (same base, same height). Thus, the area of <math>\triangle</math> <math>A'AB</math> is twice that (same height, twice the base). Similarly, [<math>\triangle</math> <math>BB'C</math>] = 2 <math>\cdot</math> [<math>\triangle</math> <math>ABC</math>], and so on.
  
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[asy]
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Adding all of these, we see that the area the four triangles around <math>ABCD</math> is twice [<math>\triangle</math> <math>DAB</math>] + [<math>\triangle</math> <math>ABC</math>] + [<math>\triangle</math> <math>BCD</math>] + [<math>\triangle</math> <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 <math>\cdot</math> [<math>ABCD</math>] = 5 <math>\cdot</math> [<math>ABCD</math>] = <math>\fbox{50}</math>.
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unitsize(1 cm);
 
  
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pair[] A, B, C, D;
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~ Mathavi
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A[0] = (0,0);
 
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B[0] = (0.6,1.2);
 
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C[0] = (-0.3,2.5);
 
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D[0] = (-1.5,0.7);
 
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B[1] = interp(A[0],B[0],2);
 
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C[1] = interp(B[0],C[0],2);
 
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D[1] = interp(C[0],D[0],2);
 
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A[1] = interp(D[0],A[0],2);
 
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draw(A[1]--B[1]--C[1]--D[1]--cycle);
 
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draw(A[0]--B[1]);
 
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draw(B[0]--C[1]);
 
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draw(C[0]--D[1]);
 
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draw(D[0]--A[1]);
 
  
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label("<math>A</math>", A[0], SW);
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Note: Anyone with a diagram would be of great help (still new to LaTex).
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label("<math>B</math>", B[0], SE);
 
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label("<math>C</math>", C[0], NE);
 
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label("<math>D</math>", D[0], NW);
 
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label("<math>A'</math>", A[1], SE);
 
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label("<math>B'</math>", B[1], NE);
 
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label("<math>C'</math>", C[1], N);
 
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label("<math>D'</math>", D[1], SW);
 
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[/asy]
 
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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.
 
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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}.
 
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~ Mathavi
 

Latest revision as of 14:24, 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

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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