2008 AMC 10B Problems/Problem 24

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Quadrilateral $ABCD$ has $AB = BC = CD$, $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$. What is the degree 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 (Cyclic Quadrilateral)

  • Note: 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 two and label the points. Create lines $\overline{AC}$ and $\overline{BD}$. We can call their 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 $BAC$ equals $55$ degrees, thus making angle $AYB$ equal to $60$ degrees. We can also find out that angle $CYB$ equals $120$ degrees.

Extend $\overline{CD}$ and $\overline{AB}$ and let their intersection be $E$. 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 $EYD$ 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.

~Minor edits by BakedPotato66

Solution 2

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,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]

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.


Solution 3(Using Trig.)

[asy] unitsize(3 cm);  pair A, B, C, D;  A = (0,0); B = dir(85); C = B + dir(-25); D = C + dir(-35); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(((A + B)/2 + scale(0.02)*rotate(90)*(B - A))--((A + B)/2 + scale(0.02)*rotate(90)*(A - B))); draw(((B + C)/2 + scale(0.02)*rotate(90)*(C - B))--((B + C)/2 + scale(0.02)*rotate(90)*(B - C))); draw(((C + D)/2 + scale(0.02)*rotate(90)*(D - C))--((C + D)/2 + scale(0.02)*rotate(90)*(C - D))); dot("$A$", A, SW); dot("$B$", B, N); dot("$C$", C, NE); dot("$D$", D, SE); label("$I$", 6/7*C); [/asy]

Let the unknown $\angle BAD$ be $x$.

First, we draw diagonal $BD$ and $AC$. $I$ is the intersection of the two diagonals. The diagonals each form two isosceles triangles, $\triangle BCD$ and $\triangle ABC$.

Using this, we find: $\angle DBC = \angle CDB = 5^\circ$ and $\angle BAC = \angle BCA = 55^\circ$. Expanding on this, we can fill in a couple more angles. $\angle ABD = 70^\circ - 5^\circ = 65^\circ$, $\angle ACD = 170^\circ - 55^\circ = 115^\circ$, $\angle BIA = \angle CID = 180^\circ - (65^\circ + 55^\circ) = 60^\circ$, $\angle BIC =   \angle AID = 180^\circ - 60^\circ = 120^\circ$.

We can rewrite $\angle CAD$ and $\angle BDA$ in terms of $x$. $\angle CAD = x - 55^\circ$ and $\angle BDA = 180^\circ - (120^\circ + x - 55^\circ) = 115^\circ - x$.

Let us relabel $AB = BC = CD = a$ and $AD = b$.

By Rule of Sines on $\triangle ACD$ and $\triangle ABD$ respectively, $\frac{\sin(\angle CAD)}{a} = \frac{\sin(\angle ACD)}{b}$, and $\frac{\sin(\angle ABD)}{b} = \frac{\sin(\angle BDA)}{a}$

In a more convenient form, $\frac{\sin(x-55^\circ)}{a} = \frac{\sin(115^\circ)}{b} \implies \frac{a}{b} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$

and $\frac{\sin(65^\circ)}{b} = \frac{\sin(115^\circ-x)}{a} \implies \frac{a}{b} = \frac{\sin(115^\circ-x)}{\sin(65^\circ)}$

$\implies \frac{\sin(115^\circ-x)}{\sin(65^\circ)} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$

Now, by identity $\sin(\theta) = \sin(180^\circ-\theta)$, $\sin(65^\circ) = \sin(115^\circ)$

Therefore, $\sin(115^\circ-x) = \sin(x-55^\circ).$ This equation is only satisfied by option $\boxed{\text {(C) } 85^\circ}$

Note: I'm pretty bad at Asymptote, if anyone could edit this and fill in the angles into the diagram, that would be pretty cool.


Solution 4 (Cheese)

Using a protractor and rule, draw an accurate diagram (Example Diagram). $\angle BAD$ looks slightly less than $90$ degrees. Therefore the answer is $\boxed{\textbf{(C) } 85}$ as $85$ is slightly less than $90$.


~Reworded by South

Solution 5 (annoying amounts of algebra + trig identities)

place A at the origin of a coordinate system, with D on the x-axis let angle BAD be $\theta$, and AB=BC=CD=1

The y value of from B-A is $\sin\theta$. The y value from C-B is $\theta-(180^\circ-70^\circ)=\theta-110^\circ$. The y value from D-C is $\theta-110^\circ-(180^\circ-170^\circ)=\theta-120^\circ$

The angles for the vectors from B to C and C to D are angle_original-(180-angle_polygon) are because the external angle of the polygon is 180-external angle, which is subtracted from the angle since it heads that amount off from the original direction.

since D-C+C-B+B-A=D-A=0 (since A, D are both on x-axis and have the same y value of 0), then: \[\sin\theta+\sin(\theta-110^\circ)+\sin(\theta-120^\circ)=0\]

from here we expand out the trig expressions using sin addition and isolate $\theta$

\[\sin\theta+\sin(\theta-110^\circ)+\sin(\theta-120^\circ)=0\] \[\sin\theta+\sin(\theta)\cos(110^\circ)-\cos(\theta)\sin(110^\circ)   +   \sin(\theta)\cos(120^\circ)-\cos(\theta)\sin(120^\circ)=0\]

\[1+\cos(110^\circ)-\cot(\theta)\sin(110^\circ)+\cos(120^\circ)-\cot(\theta)\sin(120^\circ)=0\] \[\cot(\theta)=\frac{1+\cos(110^\circ)+\cos(120^\circ)}{\sin(110^\circ)+\sin(120^\circ)}\]

At this point if you are a human calculator feel free to to solve, otherwise we want to try and evaluate the right hand side into some nice expression (ideally cot of an angle).


\[\cot(\theta)=\frac{1+\cos(120^\circ)\cos(10^\circ)+\sin(120^\circ)\sin(10^\circ)+\cos(120^\circ)} {\sin(120^\circ)\cos(10^\circ)-\cos(120^\circ)\sin(10^\circ)+\sin(120^\circ}\]

\[\cot(\theta)=\frac{1+\frac{-1}{2}\cos(10^\circ)+\frac{\sqrt{3}}{2}\sin(10^\circ)+\frac{-1}{2}} {\frac{\sqrt{3}}{2}\cos(10^\circ)-\frac{-1}{2}\sin(10^\circ)+\frac{\sqrt{3}}{2}}\]

\[\cot(\theta)=\frac{1-\cos(10^\circ)+\sqrt{3}\sin(10^\circ)} {\sqrt{3}\cos(10^\circ)+\sin(10^\circ)+\sqrt{3}}\]

since the expression still isn't simplified, notice that using the double angle identity on cosine can be used to cancel the 1, and $\sqrt{3}$

Let: $\cos(5^\circ)=a$, $\sin(5^\circ)=b$

\[\cot(\theta)=\frac{1-(1-2b^2)+\sqrt{3}\cdot(2ab)} {\sqrt{3}(2a^2-1)+2ab+\sqrt{3}}\]

\[\cot(\theta)=\frac{2b^2+2ab\sqrt{3}} {2a^2\sqrt{3}+2ab}\]

\[\cot(\theta)=\frac{2b(b+a\sqrt{3})} {2a(a\sqrt{3}+b)}\]




\[\theta=85^\circ\] ~sahan

Solution 6 (guess and check)

Obtain: $\sin\theta+\sin(\theta-110^\circ)+\sin(\theta-120^\circ)=0$ from solution 5 We now guess $\theta=85^\circ$ and try to verify

\[\sin85^\circ+\sin-25^\circ+\sin-35^\circ=0\] \[\cos5^\circ=\sin25^\circ+\sin35^\circ\] \[\cos5^\circ=\sin(30^\circ-5^\circ)+\sin(30^\circ+5^\circ)\] \[\cos5^\circ=2\sin30^\circ\cos5^\circ\] \[\cos5^\circ=\cos5^\circ\]

Solution 7 (alternate way to bash out algebra + trig identities)

Use equation from Solution 5: \[\sin\theta+\sin(\theta-110^\circ)+\sin(\theta-120^\circ)=0\] \[\sin\theta+\sin(\theta-115^\circ+5^\circ)+\sin(\theta-115^\circ-5^\circ)=0\] \[\sin\theta+2\sin(\theta-115^\circ)\cos(5^\circ)=0\] \[\sin\theta-2\sin(\theta+65^\circ)\cos(5^\circ)=0\] \[\sin\theta=2\sin(\theta+65^\circ)\cos(5^\circ)\]

now guess $\theta=85$ so that the cos(5) is dealt with (and then check it works)

If you refuse to guess work through the following algebra :D

Let: $\phi=\theta+65^\circ$

\[\sin(\phi-65^\circ)=2\sin\phi\cos(5^\circ)\] \[\sin\phi\cos(65^\circ)-\cos\phi\sin(65^\circ)=2\sin\phi\cos(5^\circ)\] \[\cos(65^\circ)-\cot(\phi)\sin(65^\circ)=2\cos(5^\circ)\] \[\cot(\phi)(\sin(60^\circ)\cos(5^\circ)+\cos(60^\circ)\sin(5^\circ))=(\cos(60^\circ)\cos(5^\circ)-\sin(60^\circ)\sin(5^\circ)) -2\cos(5^\circ)\] \[\cot(\phi)(\frac{\sqrt{3}}{2}\cos(5^\circ)+\frac{1}{2}\sin(5^\circ))=(\frac{1}{2}\cos(5^\circ)-\frac{\sqrt{3}}{2}\sin(5^\circ)) -2\cos(5^\circ)\] \[\cot(\phi)(\frac{\sqrt{3}}{2}\cos(5^\circ)+\frac{1}{2}\sin(5^\circ))=(\frac{-3}{2}\cos(5^\circ)-\frac{\sqrt{3}}{2}\sin(5^\circ))\] \[\cot(\phi)(\frac{\sqrt{3}}{2}\cos(5^\circ)+\frac{1}{2}\sin(5^\circ))=-\sqrt{3} (\frac{\sqrt{3}}{2}\cos(5^\circ)+\frac{1}{2}\sin(5^\circ))\] \[\cot(\phi)=-\sqrt3\]



\[\angle BAD = 85^\circ\]

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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