Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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Now we can compute their inner angles. <math>BP</math> is the bisector of the angle <math>ABC</math>, hence <math>\angle ABP = \angle CBP = 35^\circ</math>, and thus also <math>\angle CDP = 35^\circ</math>. <math>CP</math> is the bisector of the angle <math>BCD</math>, hence <math>\angle BCP = \angle DCP = 85^\circ</math>, and thus also <math>\angle BAP = 85^\circ</math>. | Now we can compute their inner angles. <math>BP</math> is the bisector of the angle <math>ABC</math>, hence <math>\angle ABP = \angle CBP = 35^\circ</math>, and thus also <math>\angle CDP = 35^\circ</math>. <math>CP</math> is the bisector of the angle <math>BCD</math>, hence <math>\angle BCP = \angle DCP = 85^\circ</math>, and thus also <math>\angle BAP = 85^\circ</math>. | ||
− | It follows that <math>\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ</math>. Thus the angle <math> | + | It follows that <math>\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ</math>. Thus the angle <math>APD</math> has <math>180^\circ</math>, and hence <math>P</math> does indeed lie on <math>AD</math>. Then obviously <math>\angle BAD = \angle BAP = \boxed{ 85^\circ }</math>. |
<asy> | <asy> |
Revision as of 17:00, 2 December 2013
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.