Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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==Solution 6== | ==Solution 6== | ||
+ | |||
+ | <asy> | ||
+ | unitsize(1cm); | ||
+ | defaultpen(.8); | ||
+ | real a=4; | ||
+ | pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120), E=D+a*dir(0); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(E--C); | ||
+ | draw(B--D); | ||
+ | label("$A$",A,SW); | ||
+ | label("$B$",B,SE); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,N); | ||
+ | label("$E$",P,W); | ||
+ | </asy> | ||
==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 10:56, 24 May 2020
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 .
Even Faster Solution: Above, we proved that P falls on line AD, and also , by , hence we have , which is the angle bisector of which is . Hence we have
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
This solution requires the use of cyclic quadrilateral properties but could be a bit time-consuming during the contest. To start off, draw a diagram like in solution one and label the points. Now draw the and and call this intersection point . Note that triangle is an isosceles triangle so angles and are each degrees. Since equals , angle had to equal degrees, thus making angle equal to degrees. We can also find out that angle CYB equals degrees. Extend point such that it lies on the same level of segment . Call this point . Since angle plus angle equals degrees, quadrilateral is a cyclic quadrilateral. Next, draw a line from point to point . Since angle and angle point to the same arc, angle is equal to . Since is an isosceles triangle(based on angle properties) and is also an isosceles triangle, we can find that is also an isosceles triangle. Thus, each of the other angles is degrees. Finally, we have angle equals degrees.
Solution 6
unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120), E=D+a*dir(0); draw(A--B--C--D--cycle); draw(E--C); draw(B--D); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,N); label("$E$",P,W); (Error compiling LaTeX. 6acd3e7675104c486c358123f07e94dfd1518904.asy: 14.13: no matching variable 'P')
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.