2008 AMC 10B Problems/Problem 24
Contents
Problem
Quadrilateral has
, angle
and angle
. What is the measure of angle
?
Solution
Solution 1
Draw the angle bisectors of the angles and
. These two bisectors obviously intersect. Let their intersection be
.
We will now prove that
lies on the segment
.
Note that the triangles and
are congruent, as they share the side
, and we have
and
.
Also note that for similar reasons the triangles and
are congruent.
Now we can compute their inner angles. is the bisector of the angle
, hence
, and thus also
.
is the bisector of the angle
, hence
, and thus also
.
It follows that . Thus the angle
has
, and hence
does indeed lie on
. Then obviously
.
Solution 2
Draw the diagonals and
, and suppose that they intersect at
. Then,
and
are both isosceles, so by angle-chasing, we find that
,
, and
. Draw
such that
and so that
is on
, and draw
such that
and
is on
. It follows that
and
are both equilateral. Also, it is easy to see that
and
by construction, so that
and
. Thus,
, so
is isosceles. Since
, then
, and
.
Solution 3
Again, draw the diagonals and
, and suppose that they intersect at
. We find by angle chasing the same way as in solution 2 that
and
. Applying the Law of Sines to
and
, it follows that
, so
is isosceles. We finish as we did in solution 2.
Solution 4
Start off with the same diagram as solution 1. Now draw which creates isosceles
. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that it's is
Solution 5(Cheap solution)
Draw the diagram accurately with a protractor and ruler. comes out to be
, or
. (You should always carry a protractor, ruler, and compass to every competition.)
See also
2008 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 |
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