Difference between revisions of "1964 AHSME Problems/Problem 22"
Talkinaway (talk | contribs) (Created page with "== Problem == Given parallelogram <math>ABCD</math> with <math>E</math> the midpoint of diagonal <math>BD</math>. Point <math>E</math> is connected to a point <math>F</math>...") |
Talkinaway (talk | contribs) (→Solution) |
||
Line 12: | Line 12: | ||
== Solution== | == Solution== | ||
− | If it works for a parallelogram <math>ABCD</math>, it should also work for a unit square, with <math>A(0, 0), B(0, 1), C(1, 1), D(1, 0)</math>. We are given that <math>E</math> is the midpoint of <math>BD</math>, so <math>E(0.5, 0.5)</math>. If <math>F</math> is on <math>DA</math>, then <math>F(x, 0)</math>. We note that <math>DF = 1-x</math> and <math>DA = 1</math>, so <math> | + | If it works for a parallelogram <math>ABCD</math>, it should also work for a unit square, with <math>A(0, 0), B(0, 1), C(1, 1), D(1, 0)</math>. We are given that <math>E</math> is the midpoint of <math>BD</math>, so <math>E(0.5, 0.5)</math>. If <math>F</math> is on <math>DA</math>, then <math>F(x, 0)</math>. We note that <math>DF = 1-x</math> and <math>DA = 1</math>, so <math>DF = \frac{1}{3}DA</math> means <math>1-x = \frac{1}{3}</math>, or <math>x = \frac{2}{3}</math>, and hence <math>F(\frac{2}{3}, 0)</math>. |
We note that <math>\triangle DFE</math> has a base <math>DF</math> that is <math>\frac{1}{3}</math> and an altitude from <math>E</math> to <math>DF</math> that is <math>\frac{1}{2}</math>. Therefore, <math>[DEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{12}</math>. | We note that <math>\triangle DFE</math> has a base <math>DF</math> that is <math>\frac{1}{3}</math> and an altitude from <math>E</math> to <math>DF</math> that is <math>\frac{1}{2}</math>. Therefore, <math>[DEF] = \frac{1}{2}bh = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{12}</math>. |
Revision as of 03:01, 24 July 2019
Problem
Given parallelogram with
the midpoint of diagonal
. Point
is connected to a point
in
so that
. What is the ratio of the area of
to the area of quadrilateral
?
Solution
If it works for a parallelogram , it should also work for a unit square, with
. We are given that
is the midpoint of
, so
. If
is on
, then
. We note that
and
, so
means
, or
, and hence
.
We note that has a base
that is
and an altitude from
to
that is
. Therefore,
.
Quadrilateral can be split into
and
. The first triangle is
of the unit square cut diagonally, so
. The second triangle has base
that is
and height
to
that is
. Therefore,
.
The entire quadrilateral has area
. This is
times larger than the area if
, so the ratio is
, or
.
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.