Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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− | Draw the diagonals <math>\overline{BD}</math> and <math>\overline{AC}</math>, and suppose that they intersect at <math>E</math>. Then, <math>\triangle ABC</math> and <math>\triangle BCD</math> are both isosceles, so by angle-chasing, we find that <math>\angle BAC = 55^{\circ}</math>, <math>\angle | + | Draw the diagonals <math>\overline{BD}</math> and <math>\overline{AC}</math>, and suppose that they intersect at <math>E</math>. Then, <math>\triangle ABC</math> and <math>\triangle BCD</math> are both isosceles, so by angle-chasing, we find that <math>\angle BAC = 55^{\circ}</math>, <math>\angle CBD = 5^{\circ}</math>, and <math>\angle BEA = 180 - \angle EBA - \angle BAE = 60^{\circ}</math>. Draw <math>E'</math> such that <math>EE'B = 60^{\circ}</math> and so that <math>E'</math> is on <math>\overline{AE}</math>, and draw <math>E''</math> such that <math>\angle EE''C = 60^{\circ}</math> and <math>E''</math> is on <math>\overline{DE}</math>. It follows that <math>\triangle BEE'</math> and <math>\triangle CEE''</math> are both equilateral. Also, it is easy to see that <math>\triangle BEC \cong \triangle DE''C</math> and <math>\triangle BCE \cong \triangle BAE'</math> by construction, so that <math>DE'' = BE = EE'</math> and <math>EE'' = CE = E'A</math>. Thus, <math>AE = AE' + E'E = EE'' + DE'' = DE</math>, so <math>\triangle ADE</math> is isosceles. Since <math>\angle AED = 120^{\circ}</math>, then <math>\angle DAC = \frac{180 - 120}{2} = 30^{\circ}</math>, and <math>\angle BAD = 30 + 55 = 85^{\circ}</math>. |
<asy> | <asy> | ||
import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; | import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; |
Revision as of 13:07, 26 August 2010
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 equal, as they share the side , and we have and .
Also note that for similar reasons the triangles and are equal.
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 .
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 |