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Line 3: |
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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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− |
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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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− |
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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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− |
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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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− |
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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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− |
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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);
| |
− | </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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− |
| |
− | <asy>
| |
− | unitsize(1cm);
| |
− | 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 ===
| |
− | 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
, solving into
.
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.