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

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(→‎Solution 3(Using Trig.))
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Now, by identity <math>\sin(\theta) = \sin(180^\circ-\theta)</math>, <math>\sin(65^\circ) = \sin(115^\circ)</math>
 
Now, by identity <math>\sin(\theta) = \sin(180^\circ-\theta)</math>, <math>\sin(65^\circ) = \sin(115^\circ)</math>
  
−
Therefore, <math>\sin(115^\circ-x) = \sin(x-55^\circ).</math> This equation is only satisfied by option $C.\boxed{(85^\circ)}\$
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Therefore, <math>\sin(115^\circ-x) = \sin(x-55^\circ).</math> This equation is only satisfied by option <math>C.\boxed{85^\circ}</math>
  
 
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.
 
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.

Revision as of 16:04, 20 August 2022

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 1

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 $\overline{AC}$ and $\overline{BD}$ and 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.

~Someonenumber011


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 $C.\boxed{85^\circ}$

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.

~Raghu9372

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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