Difference between revisions of "2006 AMC 10A Problems/Problem 17"
m (minor edit) |
m (→Solution 1) |
||
Line 18: | Line 18: | ||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | + | By [[symmetry]], <math>WXYZ</math> is a [[square]]. | |
<asy> | <asy> |
Revision as of 17:52, 22 October 2020
Contents
Problem
In rectangle , points
and
trisect
, and points
and
trisect
. In addition,
, and
. What is the area of quadrilateral
shown in the figure?
Solution
Solution 1
Draw . Clearly
. Then
is isosceles, and is a
. Hence
, and
.
There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.
Solution 2
Draw the lines as shown above, and count the squares. There are 12, so we have .
Solution 3
We see that if we draw a line to it is half the width of the rectangle so that length would be
, and the resulting triangle is a
so using the Pythagorean Theorem we can get that each side is
so the area of the middle square would be
which is our answer.
Solution 4
Since and
are trisection points and
, we see that
. Also,
, so triangle
is a right isosceles triangle, i.e.
. By symmetry, triangles
,
, and
are also right isosceles triangles. Therefore,
, which means triangle
is also a right isosceles triangle. Also, triangle
is a right isosceles triangle.
Then , and
. Hence,
.
By symmetry, quadrilateral is a square, so its area is
~made by AoPS (somewhere) -put here by qkddud~
See Also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.