Difference between revisions of "1980 AHSME Problems/Problem 21"
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Let <math>M</math> be the midpoint of <math>\overline{DC}</math>. Then <math>\triangle ECM \sim \triangle ACD</math> and <math>\overline{EM} || \overline{AD}</math>. Since <math>\overline{EM} || \overline{FD}</math>, it follows that <math>\triangle BFD \sim \triangle BEM</math>. Let <math>a</math> be the area of <math>\triangle BFD</math>. Since the sides of <math>\triangle BEM</math> are twice as long as the corresponding sides of <math>\triangle BFD</math>, the area of <math>\triangle BEM</math> must be <math>2^2=4</math> times the area of <math>\triangle BFD</math>, that is, <math>4a</math>. Since the height of <math>\triangle BEC</math> is the same as the height of <math>\triangle BEM</math> and the base of <math>\triangle BEC</math> is <math>\frac{3}{2}</math> times the base of <math>\triangle BEM</math>, the area of <math>\triangle BEC</math> is <math>\frac{3}{2}</math> times the area of <math>\triangle BEM</math>, or <math>\frac{3}{2} \cdot 4a = 6a</math>. Thus the area of quadrilateral <math>FDCE</math> is <math>6a - a = 5a</math>, so the ratio of the area of <math>\triangle BFD</math> to the area of quadrilateral <math>FDCE</math> is <math>\frac{a}{5a} = \frac{1}{5}</math> <math>\fbox{(A)}</math>. | Let <math>M</math> be the midpoint of <math>\overline{DC}</math>. Then <math>\triangle ECM \sim \triangle ACD</math> and <math>\overline{EM} || \overline{AD}</math>. Since <math>\overline{EM} || \overline{FD}</math>, it follows that <math>\triangle BFD \sim \triangle BEM</math>. Let <math>a</math> be the area of <math>\triangle BFD</math>. Since the sides of <math>\triangle BEM</math> are twice as long as the corresponding sides of <math>\triangle BFD</math>, the area of <math>\triangle BEM</math> must be <math>2^2=4</math> times the area of <math>\triangle BFD</math>, that is, <math>4a</math>. Since the height of <math>\triangle BEC</math> is the same as the height of <math>\triangle BEM</math> and the base of <math>\triangle BEC</math> is <math>\frac{3}{2}</math> times the base of <math>\triangle BEM</math>, the area of <math>\triangle BEC</math> is <math>\frac{3}{2}</math> times the area of <math>\triangle BEM</math>, or <math>\frac{3}{2} \cdot 4a = 6a</math>. Thus the area of quadrilateral <math>FDCE</math> is <math>6a - a = 5a</math>, so the ratio of the area of <math>\triangle BFD</math> to the area of quadrilateral <math>FDCE</math> is <math>\frac{a}{5a} = \frac{1}{5}</math> <math>\fbox{(A)}</math>. | ||
− | - | + | -j314andrews |
== Solution 2 == | == Solution 2 == |
Latest revision as of 05:24, 25 June 2025
Contents
Problem
In triangle ,
,
is the midpoint of side
,
and
is a point on side
such that
;
and
intersect at
.
The ratio of the area of triangle
to the area of quadrilateral
is
Solution 1
Let be the midpoint of
. Then
and
. Since
, it follows that
. Let
be the area of
. Since the sides of
are twice as long as the corresponding sides of
, the area of
must be
times the area of
, that is,
. Since the height of
is the same as the height of
and the base of
is
times the base of
, the area of
is
times the area of
, or
. Thus the area of quadrilateral
is
, so the ratio of the area of
to the area of quadrilateral
is
.
-j314andrews
Solution 2
We can use the principle of same height same area to solve this problem.
See also
1980 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.