Difference between revisions of "1978 AHSME Problems/Problem 29"
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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 \cdot [<math>\triangle</math> <math>ABC</math>], and so on. | + | 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. |
− | 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 \cdot [<math>ABCD</math>] = 5 \cdot [<math>ABCD</math>] = \fbox{50}. | + | 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>. |
~ Mathavi | ~ Mathavi | ||
Note: Anyone with a diagram would be of great help (still new to LaTex). | Note: Anyone with a diagram would be of great help (still new to LaTex). |
Revision as of 14:04, 23 August 2022
Problem
Sides and
, respectively, of convex quadrilateral
are extended past
and
to points
and
. Also,
and
; and the area of
is
. The area of
is
Solution
Notice that the area of
is the same as that of
(same base, same height). Thus, the area of
is twice that (same height, twice the base). Similarly, [
] = 2
[
], and so on.
Adding all of these, we see that the area the four triangles around is twice [
] + [
] + [
] + [
], which is itself twice the area of the quadrilateral
. Finally, [
] = [
] + 4
[
] = 5
[
] =
.
~ Mathavi
Note: Anyone with a diagram would be of great help (still new to LaTex).