Difference between revisions of "1992 AHSME Problems/Problem 24"
m (Added a little more detail to the solution) |
(Added 2nd solution) |
||
Line 9: | Line 9: | ||
\text{(E) } 6</math> | \text{(E) } 6</math> | ||
− | == Solution == | + | == Solution 1 == |
<math>\fbox{C}</math> Use vectors. Place an origin at <math>A</math>, with <math>B = p, D = q, C = p + q</math>. We know that <math>\|p \times q\|=10</math>, and also <math>E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q</math>, and now we can find the area of <math>EFHG</math> by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero). | <math>\fbox{C}</math> Use vectors. Place an origin at <math>A</math>, with <math>B = p, D = q, C = p + q</math>. We know that <math>\|p \times q\|=10</math>, and also <math>E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q</math>, and now we can find the area of <math>EFHG</math> by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero). | ||
+ | == Solution 2 == | ||
+ | By noting the following: <cmath>\begin{align*}|EFHG| &= |\triangle EGH| + |\triangle EFH| \\ &= \frac{1}{2}|AEHD|+\frac{1}{2}|EHCB| \\ &= \frac{1}{2}[|AEHD|+EHCB|] \\ &= \frac{1}{2}|ABCD| \\ &= 5 \end{align*}</cmath> we see that the answer is <math>\fbox{C}</math>. | ||
== See also == | == See also == | ||
{{AHSME box|year=1992|num-b=23|num-a=25}} | {{AHSME box|year=1992|num-b=23|num-a=25}} |
Revision as of 07:29, 28 June 2018
Contents
Problem
Let be a parallelogram of area
with
and
. Locate
and
on segments
and
, respectively, with
. Let the line through
parallel to
intersect
at
. The area of quadrilateral
is
Solution 1
Use vectors. Place an origin at
, with
. We know that
, and also
, and now we can find the area of
by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero).
Solution 2
By noting the following: we see that the answer is
.
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.