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| | <math>\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95</math> | | <math>\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95</math> |
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| − | ==Solution==
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| − | ===Solution 1 ===
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| − | 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>.
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| − | We will now prove that <math>P</math> lies on the segment <math>AD</math>.
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| − | Note that the triangles <math>ABP</math> and <math>CBP</math> are congruent, as they share the side <math>BP</math>, and we have <math>AB=BC</math> and <math>\angle ABP = \angle CBP</math>.
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| − | Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are congruent.
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| − | 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>. (Faster Solution picks up here) <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>.
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| − | It follows that <math>\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ</math>. Thus the angle <math>APD</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>.
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| − |
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| − | <asy>
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| − | unitsize(1cm);
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| − | defaultpen(.8);
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| − | real a=4;
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| − | pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);
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| − | draw(A--B--C--D--cycle);
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| − | pair P1=B+3*a*dir(145), P2=C+3*a*dir(205);
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| − | pair P=intersectionpoint(B--P1,C--P2);
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| − | draw(B--P--C);
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| − | label("$A$",A,SW);
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| − | label("$B$",B,SE);
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| − | label("$C$",C,NE);
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| − | label("$D$",D,N);
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| − | label("$P$",P,W);
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| − |
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| − | label("$35^\circ$",B + dir(180-17.5));
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| − | label("$35^\circ$",B + dir(180-35-17.5));
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| − |
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| − | label("$85^\circ$",C + .5*dir(120+42.5));
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| − | label("$85^\circ$",C + .5*dir(120+85+42.5));
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| − | </asy>
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| − |
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| − | Faster Solution: Because we now know three angles, we can subtract to get <math>360 - 35 - 85 - 85 - 35 - 35</math>, or <math>\boxed{85}</math>.
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| − | Even Faster Solution: Above, we proved that P falls on line AD, and also <math>\triangle ABP = \triangle CBP</math>, by <math>SAS</math>, hence we have <math>\angle BCP=\angle BAP</math>, which is the angle bisector of <math>\angle BCD</math> which is <math>\dfrac{170}{2}=85</math>. Hence we have <math>\angle BCP=\angle BAP=\angle BAD= 85^\circ</math>
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| − |
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| − | === Solution 2 ===
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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 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>.
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| − | <asy>
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| − | 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;
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| − | pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0);
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| − | 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));
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| − | 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);
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| − | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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| − | </asy>
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| − |
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| − | ===Solution 3 ===
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| − | Again, draw the diagonals <math>\overline{BD}</math> and <math>\overline{AC}</math>, and suppose that they intersect at <math>E</math>. We find by angle chasing the same way as in solution 2 that <math>m\angle ABE = 65^\circ</math> and <math>m\angle DCE = 115^\circ</math>. Applying the Law of Sines to <math>\triangle AEB</math> and <math>\triangle EDC</math>, it follows that <math>DE = \frac{2\sin 115^\circ}{\sin \angle DEC} = \frac{2\sin 65^\circ}{\sin \angle AEB} = EA</math>, so <math>\triangle AED</math> is isosceles. We finish as we did in solution 2.
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| − |
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| − | <asy>
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| − | unitsize(1cm);
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| − | defaultpen(.8);
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| − | real a=4;
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| − | pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);
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| − | draw(A--B--C--D--cycle);
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| − | pair P=intersectionpoint(B--D,C--A);
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| − | draw(A--C); draw(B--D);
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| − | label("$A$",A,SW);
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| − | label("$B$",B,SE);
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| − | label("$C$",C,NE);
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| − | label("$D$",D,N);
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| − | label("$E$",P,W);
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| − | </asy>
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| − |
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| − | === Solution 4 ===
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| − | Start off with the same diagram as solution 1. Now draw <math>\overline{CA}</math> which creates isosceles <math>\triangle CAB</math>. 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 <math>\boxed{85}.</math>
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| | | | |
| | ==Solution 5== | | ==Solution 5== |
Problem
Quadrilateral 
has 
, angle 
and angle 
. What is the measure of angle 
?
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
First, connect the diagonal 
, then, draw line 
such that it is congruent to 
and is parallel to . Because triangle 
is isosceles and angle is 
, the angles and are both 
. Because angle 
is 
, we get angle 
is 
. Next, noticing parallel lines and and transversal , we see that angle 
is also , and subtracting off angle gives that angle 
is 
.
Now, because we drew 
, triangle 
is equilateral. We can also conclude that 
meaning that triangle 
is isosceles, and angles 
and 
are equal.
Finally, we can set up our equation. Denote angle as 
. Then, because 
is a parallelogram, the angle 
is also . Then, is 
. Again because is a parallelogram, angle 
is 
. Subtracting angle gives that angle equals 
. Because angle equals angle , we get ![\[x-60=110-x\]](//latex.artofproblemsolving.com/0/1/f/01fa4b5d2675f2c894d513b672bf345b5a788893.png)
, solving into 
.
![[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); draw(B--E); draw(D--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,SE); label("$D$",D,N); label("$E$",E,NE); label("$60^\circ$",C + .75*dir(360-65-115-55-30)); label("$65^\circ$",B + .75*dir(180-32.5)); label("$x^\circ$",A + .5*dir(42.5)); label("$5^\circ$",D + 2.5*dir(360-60-2.5)); label("$60^\circ$",D + .75*dir(360-30)); label("$60^\circ$",E + .5*dir(360-150)); label("$5^\circ$",B + 2.5*dir(180-65-2.5)); [/asy]](//latex.artofproblemsolving.com/3/b/e/3be2a3a08f00b0d4c6626443c6e9c42bc115eb5f.png)
Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles.
~Someonenumber011
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
has 
, angle 
and angle 
. What is the measure of angle 
?
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 
. Since angle 
plus angle 
equals 
degrees, quadrilateral 
is a cyclic quadrilateral. Next, draw a line from point 
and angle 
point to the same arc, angle 
. 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 
degrees.
, then, draw line 
such that it is congruent to 
and is parallel to 
is isosceles and angle 
, the angles 
. Because angle 
is 
, we get angle 
is 
. Next, noticing parallel lines 
is also 
is 
.
, triangle 
is equilateral. We can also conclude that 
meaning that triangle 
is isosceles, and angles 
and 
are equal.
. Then, because 
is a parallelogram, the angle 
is also 
. Again because 
is 
. Subtracting angle 
. Because angle ![\[x-60=110-x\]](http://latex.artofproblemsolving.com/0/1/f/01fa4b5d2675f2c894d513b672bf345b5a788893.png)
, solving into 
.
![[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); draw(B--E); draw(D--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,SE); label("$D$",D,N); label("$E$",E,NE); label("$60^\circ$",C + .75*dir(360-65-115-55-30)); label("$65^\circ$",B + .75*dir(180-32.5)); label("$x^\circ$",A + .5*dir(42.5)); label("$5^\circ$",D + 2.5*dir(360-60-2.5)); label("$60^\circ$",D + .75*dir(360-30)); label("$60^\circ$",E + .5*dir(360-150)); label("$5^\circ$",B + 2.5*dir(180-65-2.5)); [/asy]](http://latex.artofproblemsolving.com/3/b/e/3be2a3a08f00b0d4c6626443c6e9c42bc115eb5f.png)
