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Difference between revisions of "2008 AMC 10B Problems/Problem 24"

m (→‎Problem)
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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);
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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 $ABCD$ has $AB = BC = CD$, angle $ABC = 70$ and angle $BCD = 170$. What is the measure of angle $BAD$?

$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$

Solution

Draw the angle bisectors of the angles $ABC$ and $BCD$. These two bisectors obviously intersect. Let their intersection be $P$. We will now prove that $P$ lies on the segment $AD$.

Note that the triangles $ABP$ and $CBP$ are equal, as they share the side $BP$, and we have $AB=BC$ and $\angle ABP = \angle CBP$.

Also note that for similar reasons the triangles $CBP$ and $CDP$ are equal.

Now we can compute their inner angles. $BP$ is the bisector of the angle $ABC$, hence $\angle ABP = \angle CBP = 35^\circ$, and thus also $\angle CDP = 35^\circ$. $CP$ is the bisector of the angle $BCD$, hence $\angle BCP = \angle DCP = 85^\circ$, and thus also $\angle BAP = 85^\circ$.

It follows that $\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ$. Thus the angle $APB$ has $180^\circ$, and hence $P$ does indeed lie on $AD$. Then obviously $\angle BAD = \angle BAP = \boxed{ 85^\circ }$.

[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

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
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