Difference between revisions of "2008 AMC 10B Problems/Problem 24"
m (→Solution 4: there was a typo where a point 'E' is used even though it did not exist, and point 'C' was supposed to be used) |
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=== Solution 4 === | === Solution 4 === | ||
− | Start off with the same diagram as solution 1. Now draw <math>\overline{CA}</math> which creates isosceles <math>\triangle CAB</math>. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that | + | Start off with the same diagram as solution 1. Now draw <math>\overline{CA}</math> which creates isosceles <math>\triangle CAB</math>. 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 <math>\boxed{85}.</math> |
+ | |||
+ | ==Solution 5(Cheap solution)== | ||
+ | Draw the diagram accurately with a protractor and ruler. <math>\angle BAD</math> comes out to be <math>\boxed{85}, or </math>\boxed{\text{C}}$. (You should always carry a protractor, ruler, and compass to every competition.) | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=23|num-a=25}} | {{AMC10 box|year=2008|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:43, 1 October 2017
Contents
Problem
Quadrilateral has , angle and angle . What is the measure of angle ?
Solution
Solution 1
Draw the angle bisectors of the angles and . These two bisectors obviously intersect. Let their intersection be . We will now prove that lies on the segment .
Note that the triangles and are congruent, as they share the side , and we have and .
Also note that for similar reasons the triangles and are congruent.
Now we can compute their inner angles. is the bisector of the angle , hence , and thus also . is the bisector of the angle , hence , and thus also .
It follows that . Thus the angle has , and hence does indeed lie on . Then obviously .
Solution 2
Draw the diagonals and , and suppose that they intersect at . Then, and are both isosceles, so by angle-chasing, we find that , , and . Draw such that and so that is on , and draw such that and is on . It follows that and are both equilateral. Also, it is easy to see that and by construction, so that and . Thus, , so is isosceles. Since , then , and .
Solution 3
Again, draw the diagonals and , and suppose that they intersect at . We find by angle chasing the same way as in solution 2 that and . Applying the Law of Sines to and , it follows that , so is isosceles. We finish as we did in solution 2.
Solution 4
Start off with the same diagram as solution 1. Now draw which creates isosceles . We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that it's is
Solution 5(Cheap solution)
Draw the diagram accurately with a protractor and ruler. comes out to be \boxed{\text{C}}$. (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.