Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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==Solution== | ==Solution== | ||
− | + | ===Solution 1 === | |
Draw the angle bisectors of the angles <math>ABC</math> and <math>BCD</math>. These two bisectors obviously intersect. Let their intersection be <math>P</math>. | Draw the angle bisectors of the angles <math>ABC</math> and <math>BCD</math>. These two bisectors obviously intersect. Let their intersection be <math>P</math>. | ||
We will now prove that <math>P</math> lies on the segment <math>AD</math>. | We will now prove that <math>P</math> lies on the segment <math>AD</math>. | ||
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label("$85^\circ$",C + .5*dir(120+42.5)); | label("$85^\circ$",C + .5*dir(120+42.5)); | ||
label("$85^\circ$",C + .5*dir(120+85+42.5)); | label("$85^\circ$",C + .5*dir(120+85+42.5)); | ||
+ | </asy> | ||
+ | |||
+ | === Solution 2 === | ||
+ | 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 </math>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}$. | ||
+ | <asy> | ||
+ | import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; | ||
+ | pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); | ||
+ | filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); | ||
+ | dot((0,0),ds); label("$A$",(-0.096,0.005),NE*lsf); dot((1,0),ds); label("$B$",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label("$C$",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label("$D$",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label("$E$",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label("$E'$",(0.1,0.23),NE*lsf); label("$60^\circ$",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label("$E''$",(0.423,0.957),NE*lsf); label("$60^\circ$",(0.761,0.886),NE*lsf,qqwuqq); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
</asy> | </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}} |
Revision as of 12:06, 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 , CBD = 5^{\circ}\angle BEA = 180 - \angle EBA - \angle BAE = 60^{\circ}E'EE'B = 60^{\circ}E'\overline{AE}E\angle EEC = 60^{\circ}E\overline{DE}\triangle BEE'\triangle CEE\triangle BEC \cong \triangle DEC\triangle BCE \cong \triangle BAE'DE = BE = EE'EE = CE = E'AAE = AE' + E'E = EE + DE = DE\triangle ADE\angle AED = 120^{\circ}\angle DAC = \frac{180 - 120}{2} = 30^{\circ}\angle BAD = 30 + 55 = 85^{\circ}$.
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 |