Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are congruent. | Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are congruent. | ||
− | 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>. (Faster Solution picks up here) <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>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>. | 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>. | ||
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label("$85^\circ$",C + .5*dir(120+85+42.5)); | label("$85^\circ$",C + .5*dir(120+85+42.5)); | ||
</asy> | </asy> | ||
+ | |||
+ | Faster Solution: Because we now know three angles, we can subtract to get <math>360 - 35 - 85 - 85 - 35 - 35</math>, or <math>\boxed{85}</math>. | ||
=== Solution 2 === | === Solution 2 === |
Revision as of 22:43, 12 February 2019
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 . (Faster Solution picks up here) 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 .
Faster Solution: Because we now know three angles, we can subtract to get , or .
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 and ruler to a 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.