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

(Solution 6)
(Solution 6)
Line 104: Line 104:
label("$60^\circ$",C + .5*dir(360-65-115-55-30));
label("$60^\circ$",C + .5*dir(360-65-115-55-30));
label("$65^\circ$",B + .5*dir(180-32.5));
label("$65^\circ$",B + .75*dir(180-32.5));
label("$x^\circ$",A + .5*dir(42.5));
label("$x^\circ$",A + .5*dir(42.5));
label("$5^\circ$",D + 2.5*dir(360-60-2.5));
label("$5^\circ$",D + 2.5*dir(360-60-2.5));
label("$60^\circ$",D + .5*dir(360-30));
label("$60^\circ$",D + .75*dir(360-30));
label("$60^\circ$",E + .5*dir(360-150));
label("$60^\circ$",E + .5*dir(360-150));
label("$5^\circ$",B + 2.5*dir(180-65-2.5));
label("$5^\circ$",B + 2.5*dir(180-65-2.5));

Revision as of 12:46, 24 May 2020


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 1

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 congruent, 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 congruent.

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 $\angle CDP = 35^\circ$. (Faster Solution picks up here) $CP$ 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 $APD$ 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]

Faster Solution: Because we now know three angles, we can subtract to get $360 - 35 - 85 - 85 - 35 - 35$, or $\boxed{85}$.

Even Faster Solution: Above, we proved that P falls on line AD, and also $\triangle ABP = \triangle CBP$, by $SAS$, hence we have $\angle BCP=\angle BAP$, which is the angle bisector of $\angle BCD$ which is $\dfrac{170}{2}=85$. Hence we have $\angle BCP=\angle BAP=\angle BAD= 85^\circ$

Solution 2

Draw the diagonals $\overline{BD}$ and $\overline{AC}$, and suppose that they intersect at $E$. Then, $\triangle ABC$ and $\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\angle BAC = 55^{\circ}$, $\angle CBD = 5^{\circ}$, and $\angle BEA = 180 - \angle EBA - \angle BAE = 60^{\circ}$. Draw $E'$ such that $EE'B = 60^{\circ}$ and so that $E'$ is on $\overline{AE}$, and draw $E''$ such that $\angle EE''C = 60^{\circ}$ and $E''$ is on $\overline{DE}$. It follows that $\triangle BEE'$ and $\triangle CEE''$ are both equilateral. Also, it is easy to see that $\triangle BEC \cong \triangle DE''C$ and $\triangle BCE \cong \triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\triangle ADE$ is isosceles. Since $\angle AED = 120^{\circ}$, then $\angle DAC = \frac{180 - 120}{2} = 30^{\circ}$, and $\angle BAD = 30 + 55 = 85^{\circ}$. [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]

Solution 3

Again, draw the diagonals $\overline{BD}$ and $\overline{AC}$, and suppose that they intersect at $E$. We find by angle chasing the same way as in solution 2 that $m\angle ABE = 65^\circ$ and $m\angle DCE = 115^\circ$. Applying the Law of Sines to $\triangle AEB$ and $\triangle EDC$, it follows that $DE = \frac{2\sin 115^\circ}{\sin \angle DEC} = \frac{2\sin 65^\circ}{\sin \angle AEB} = EA$, so $\triangle AED$ is isosceles. We finish as we did in solution 2.

[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 P=intersectionpoint(B--D,C--A); draw(A--C); draw(B--D); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,N); label("$E$",P,W); [/asy]

Solution 4

Start off with the same diagram as solution 1. Now draw $\overline{CA}$ which creates isosceles $\triangle CAB$. 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 $\boxed{85}.$

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 $\overline{AC}$ and $\overline{BD}$ and call this intersection point $Y$. Note that triangle $BCD$ is an isosceles triangle so angles $CDB$ and $CBD$ are each $5$ degrees. Since $AB$ equals $BC$, angle $BAC$ had to equal $55$ degrees, thus making angle $AYB$ equal to $60$ degrees. We can also find out that angle CYB equals $120$ degrees. Extend point $C$ such that it lies on the same level of segment $AB$. Call this point $E$. Since angle $BEC$ plus angle $CYB$ equals $180$ degrees, quadrilateral $YCEB$ is a cyclic quadrilateral. Next, draw a line from point $Y$ to point $E$. Since angle $YBC$ and angle $YEC$ point to the same arc, angle $YEC$ is equal to $5 degrees$. Since $EBD$ is an isosceles triangle(based on angle properties) and $YAE$ is also an isosceles triangle, we can find that $YAD$ is also an isosceles triangle. Thus, each of the other angles is $\frac{180-120}{2}=30$ degrees. Finally, we have angle $BAD$ equals $30+55=\boxed{85}$ degrees.

Solution 6

First, connect the diagonal $DB$, then, draw line $DE$ such that it is congruent to $DC$ and is parallel to $AB$. Because triangle $DCB$ is isosceles and angle $DCB$ is $170^\circ$, the angles $CDB$ and $CBD$ are both $\frac{180-170}{2} = 5^\circ$. Because angle $ABC$ is $70^\circ$, we get angle $ABD$ is $65^\circ$. Next, noticing parallel lines $AB$ and $DE$ and transversal $DB$, we see that angle $BDE$ is also $65^\circ$, and subtracting off angle $CDB$ gives that angle $EDC$ is $60^\circ$.

Now, because we drew $ED = DC$, triangle $DEC$ is equilateral. We can also conclude that $EC=DC=CB$ meaning that triangle $ECB$ is isosceles, and angles $CBE$ and $CEB$ are equal.

Finally, we can set up our equation. Denote angle $BAD$ as $x^\circ$. Then, because $ABED$ is a parallelogram, the angle $DEB$ is also $x^\circ$. Then, $CEB$ is $(x-60)^\circ$. Again because $ABED$ is a parallelogram, angle $ABE$ is $(180-x)^\circ$. Subtracting angle $ABC$ gives that angle $CBE$ equals $(110-x)^\circ$. Because angle $CBE$ equals angle $CEB$, we get \[x-60=110-x\], solving into $x=\boxed{85^\circ}$.

[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), 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,NE); label("$D$",D,N); label("$E$",E,NE); label("$60^\circ$",C + .5*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]

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.


See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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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