Difference between revisions of "2006 AMC 10B Problems/Problem 24"
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== Solution 1 (Similar Triangles) == | == Solution 1 (Similar Triangles) == | ||
− | When we see this problem, it practically screams similar triangles at us. Extend <math>OP</math> to the left until it intersects lines <math>AD</math> and <math>BC</math> at point <math>E</math>. Triangles <math> | + | When we see this problem, it practically screams similar triangles at us. Extend <math>OP</math> to the left until it intersects lines <math>AD</math> and <math>BC</math> at point <math>E</math>. Triangles <math>EBO</math> and <math>ECP</math> are similar, and by symmetry, so are triangles <math>EAO</math> and <math>EDP</math>. Then, the area of kite <math>EAOB</math> is just <math>4 \sqrt{2} \times 2</math> and the area of kite <math>EDPC</math> is <math>8 \sqrt{2} \times 4</math> (using our similarity ratios). The difference of these yields the area of hexagon <math>AOBCPD</math> or <math>24\sqrt{2} \Longrightarrow \boxed{\mathrm{(B)}}</math>. |
~icecreamrolls8 | ~icecreamrolls8 | ||
Line 52: | Line 52: | ||
Note: Quadrilaterals <math>OADP</math> and <math>OBCP</math> are congruent, so they have equal areas. | Note: Quadrilaterals <math>OADP</math> and <math>OBCP</math> are congruent, so they have equal areas. | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://www.youtube.com/watch?v=EmUfNAxLZ7A ~David | ||
== See also == | == See also == |
Latest revision as of 14:38, 6 April 2024
Contents
Problem
Circles with centers and
have radii
and
, respectively, and are externally tangent. Points
and
on the circle with center
and points
and
on the circle with center
are such that
and
are common external tangents to the circles. What is the area of the concave hexagon
?
Video Solution 1
~ Education, the Study of Everything
Solution 1 (Similar Triangles)
When we see this problem, it practically screams similar triangles at us. Extend to the left until it intersects lines
and
at point
. Triangles
and
are similar, and by symmetry, so are triangles
and
. Then, the area of kite
is just
and the area of kite
is
(using our similarity ratios). The difference of these yields the area of hexagon
or
.
~icecreamrolls8
Solution 2 (Perpendiculars)
File is too big, so go to https://www.imgur.com/a/7aphGaa
Sorry for the wrong point names, I didn't know how to change them.
Since a tangent line is perpendicular to the radius containing the point of tangency, .
Construct a perpendicular to that goes through point
. Label the point of intersection
.
Clearly is a rectangle, so
and
. By the Pythagorean Theorem,
.
The area of is
. The area of
is
, so the area of quadrilateral
is
. Using similar steps, the area of quadrilateral
is also
. Therefore, the area of hexagon
is
.
Note: Quadrilaterals and
are congruent, so they have equal areas.
Video Solution
https://www.youtube.com/watch?v=EmUfNAxLZ7A ~David
See also
2006 AMC 10B (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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.