Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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==Solution== | ==Solution== | ||
+ | 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>. | ||
+ | |||
+ | Note that the triangles <math>ABP</math> and <math>CBP</math> are equal, as they share the side <math>BP</math>, and we have <math>AB=BC</math> and <math>\angle ABP = \angle CBP</math>. | ||
+ | |||
+ | Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are equal. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | It follows that <math>\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ</math>. Thus the angle <math>APB</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>. | ||
+ | |||
+ | <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); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); | ||
+ | pair P=intersectionpoint(B--P1,C--P2); | ||
+ | draw(B--P--C); | ||
+ | label("$A$",A,SW); | ||
+ | label("$B$",B,SE); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,N); | ||
+ | label("$P$",P,W); | ||
+ | |||
+ | label("$35^\circ$",B + dir(180-17.5)); | ||
+ | label("$35^\circ$",B + dir(180-35-17.5)); | ||
+ | |||
+ | label("$85^\circ$",C + .5*dir(120+42.5)); | ||
+ | label("$85^\circ$",C + .5*dir(120+85+42.5)); | ||
+ | </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 15:14, 29 March 2009
Problem
Quadrilateral has , angle and angle . What is the measure of angle ?
Solution
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 .
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 |