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

(→‎Solution 6)
(→‎Solution 2)
 
(48 intermediate revisions by 18 users not shown)
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==Problem==
 
==Problem==
−
Quadrilateral <math>ABCD</math> has <math>AB = BC = CD</math>, angle <math>ABC = 70</math> and angle <math>BCD = 170</math>. What is the measure of angle <math>BAD</math>?
+
Quadrilateral <math>ABCD</math> has <math>AB = BC = CD</math>, <math>m\angle ABC = 70^\circ</math> and <math>m\angle BCD = 170^\circ</math>. What is the degree measure of <math>\angle BAD</math>?
  
 
<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>
  
−
==Solution==
+
==Solution 1 (Cyclic Quadrilateral)==
−
===Solution 1 ===
+
* Note: This solution requires the use of cyclic quadrilateral properties but could be a bit time-consuming during the contest.
−
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>.
 
−
We will now prove that <math>P</math> lies on the segment <math>AD</math>.  
 
  
−
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>.
+
To start off, draw a diagram like in solution two and label the points. Create lines <math>\overline{AC}</math> and <math>\overline{BD}</math>. We can call their intersection point <math>Y</math>. Note that triangle <math>BCD</math> is an isosceles triangle so angles <math>CDB</math> and <math>CBD</math> are each <math>5</math> degrees. Since <math>AB</math> equals <math>BC</math>, angle <math>BAC</math> equals <math>55</math> degrees, thus making angle <math>AYB</math> equal to <math>60</math> degrees. We can also find out that angle <math>CYB</math> equals <math>120</math> degrees.  
  
−
Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are congruent.
+
Extend <math>\overline{CD}</math> and <math>\overline{AB}</math> and let their intersection be <math>E</math>. Since angle <math>BEC</math> plus angle <math>CYB</math> equals <math>180</math> degrees, quadrilateral <math>YCEB</math> is a cyclic quadrilateral.  
  
−
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>.
+
Next, draw a line from point <math>Y</math> to point <math>E</math>. Since angle <math>YBC</math> and angle <math>YEC</math> point to the same arc, angle <math>YEC</math> is equal to <math>5</math> degrees. Since <math>EYD</math> is an isosceles triangle (based on angle properties) and <math>YAE</math> is also an isosceles triangle, we can find that <math>YAD</math> is also an isosceles triangle. Thus, each of the other angles is <math>\frac{180-120}{2}=30</math> degrees. Finally, we have angle <math>BAD</math> equals <math>30+55=\boxed{85}</math> degrees.
 +
 
 +
~Minor edits by BakedPotato66
 +
 
 +
==Solution 2==
 +
 
 +
First, connect the diagonal <math>DB</math>, then, draw line <math>DE</math> such that it is congruent to <math>DC</math> and is parallel to <math>AB</math>. Because triangle <math>DCB</math> is isosceles and angle <math>DCB</math> is <math>170^\circ</math>, the angles <math>CDB</math> and <math>CBD</math> are both <math>\frac{180-170}{2} = 5^\circ</math>. Because angle <math>ABC</math> is <math>70^\circ</math>, we get angle <math>ABD</math> is <math>65^\circ</math>. Next, noticing parallel lines <math>AB</math> and <math>DE</math> and transversal <math>DB</math>, we see that angle <math>BDE</math> is also <math>65^\circ</math>, and subtracting off angle <math>CDB</math> gives that angle <math>EDC</math> is <math>60^\circ</math>.
 +
 
 +
Now, because we drew <math>ED = DC</math>, triangle <math>DEC</math> is equilateral. We can also conclude that <math>EC=DC=CB</math> meaning that triangle <math>ECB</math> is isosceles, and angles <math>CBE</math> and <math>CEB</math> are equal.
 +
 
 +
Finally, we can set up our equation. Denote angle <math>BAD</math> as <math>x^\circ</math>. Then, because <math>ABED</math> is a parallelogram, the angle <math>DEB</math> is also <math>x^\circ</math>. Then, <math>CEB</math> is <math>(x-60)^\circ</math>. Again because <math>ABED</math> is a parallelogram, angle <math>ABE</math> is <math>(180-x)^\circ</math>. Subtracting angle <math>ABC</math> gives that angle <math>CBE</math> equals <math>(110-x)^\circ</math>. Because angle <math>CBE</math> equals angle <math>CEB</math>, we get <cmath>x-60=110-x</cmath>, solving into <math>x=\boxed{85^\circ}</math>.
  
−
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>.
 
  
 
<asy>
 
<asy>
Line 21: Line 28:
 
defaultpen(.8);
 
defaultpen(.8);
 
real a=4;
 
real a=4;
−
pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);
+
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(A--B--C--D--cycle);
−
pair P1=B+3*a*dir(145), P2=C+3*a*dir(205);
+
draw(E--C);
−
pair P=intersectionpoint(B--P1,C--P2);
+
draw(B--D);
−
draw(B--P--C);
+
draw(B--E);
 +
draw(D--E);
 
label("$A$",A,SW);
 
label("$A$",A,SW);
 
label("$B$",B,SE);
 
label("$B$",B,SE);
−
label("$C$",C,NE);
+
label("$C$",C,SE);
 
label("$D$",D,N);
 
label("$D$",D,N);
−
label("$P$",P,W);
+
label("$E$",E,NE);
−
 
+
label("$60^\circ$",C + .75*dir(360-65-115-55-30));
−
label("$35^\circ$",B + dir(180-17.5));
+
label("$65^\circ$",B + .75*dir(180-32.5));
−
label("$35^\circ$",B + dir(180-35-17.5));
+
label("$x^\circ$",A + .5*dir(42.5));
−
 
+
label("$5^\circ$",D + 2.5*dir(360-60-2.5));
−
label("$85^\circ$",C + .5*dir(120+42.5));
+
label("$60^\circ$",D + .75*dir(360-30));
−
label("$85^\circ$",C + .5*dir(120+85+42.5));
+
label("$60^\circ$",E + .5*dir(360-150));
 +
label("$5^\circ$",B + 2.5*dir(180-65-2.5));
 
</asy>
 
</asy>
  
−
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>.
+
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
  
−
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>
+
==Solution 3(Using Trig.)==
  
−
=== Solution 2 ===
 
−
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>.
 
 
<asy>
 
<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;
+
unitsize(3 cm);
−
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 ===
+
pair A, B, C, D;
−
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.
 
  
−
<asy>
+
A = (0,0);
−
unitsize(1cm);
+
B = dir(85);
−
defaultpen(.8);
+
C = B + dir(-25);
−
real a=4;
+
D = C + dir(-35);
−
pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120);
 
 
draw(A--B--C--D--cycle);
 
draw(A--B--C--D--cycle);
−
pair P=intersectionpoint(B--D,C--A);
+
draw(A--C);
−
draw(A--C); draw(B--D);
+
draw(B--D);
−
label("$A$",A,SW);
+
draw(((A + B)/2 + scale(0.02)*rotate(90)*(B - A))--((A + B)/2 + scale(0.02)*rotate(90)*(A - B)));
−
label("$B$",B,SE);
+
draw(((B + C)/2 + scale(0.02)*rotate(90)*(C - B))--((B + C)/2 + scale(0.02)*rotate(90)*(B - C)));
−
label("$C$",C,NE);
+
draw(((C + D)/2 + scale(0.02)*rotate(90)*(D - C))--((C + D)/2 + scale(0.02)*rotate(90)*(C - D)));
−
label("$D$",D,N);
+
dot("$A$", A, SW);
−
label("$E$",P,W);
+
dot("$B$", B, N);
 +
dot("$C$", C, NE);
 +
dot("$D$", D, SE);
 +
label("$I$", 6/7*C);
 
</asy>
 
</asy>
  
−
=== Solution 4 ===
+
Let the unknown <math>\angle BAD</math> be <math>x</math>.
−
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>
+
 
 +
First, we draw diagonal <math>BD</math> and <math>AC</math>.
 +
<math>I</math> is the intersection of the two diagonals. The diagonals each form two isosceles triangles, <math>\triangle BCD</math> and <math>\triangle ABC</math>.
 +
 
 +
Using this, we find: <math>\angle DBC = \angle CDB = 5^\circ</math> and <math>\angle BAC = \angle BCA = 55^\circ</math>. Expanding on this, we can fill in a couple more angles.
 +
<math>\angle ABD = 70^\circ - 5^\circ = 65^\circ</math>, <math>\angle ACD = 170^\circ - 55^\circ = 115^\circ</math>, <math>\angle BIA = \angle CID = 180^\circ - (65^\circ + 55^\circ) = 60^\circ</math>, <math>\angle BIC =  
 +
\angle AID = 180^\circ - 60^\circ = 120^\circ</math>.
 +
 
 +
We can rewrite <math>\angle CAD</math> and <math>\angle BDA</math> in terms of <math>x</math>. <math>\angle CAD = x - 55^\circ</math> and <math>\angle BDA = 180^\circ - (120^\circ + x - 55^\circ) = 115^\circ - x</math>.
 +
 
 +
Let us relabel <math>AB = BC = CD = a</math> and <math>AD = b</math>.
 +
 
 +
By Rule of Sines on <math>\triangle ACD</math> and <math>\triangle ABD</math> respectively, <math>\frac{\sin(\angle CAD)}{a} = \frac{\sin(\angle ACD)}{b}</math>, and <math>\frac{\sin(\angle ABD)}{b} = \frac{\sin(\angle BDA)}{a}</math>
 +
 
 +
In a more convenient form, <math>\frac{\sin(x-55^\circ)}{a} = \frac{\sin(115^\circ)}{b} \implies \frac{a}{b} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}</math>
 +
 
 +
and <math>\frac{\sin(65^\circ)}{b} = \frac{\sin(115^\circ-x)}{a} \implies \frac{a}{b} = \frac{\sin(115^\circ-x)}{\sin(65^\circ)}</math>
  
−
==Solution 5==
+
<math>\implies \frac{\sin(115^\circ-x)}{\sin(65^\circ)} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}</math>
−
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 <math>\overline{AC}</math> and <math>\overline{BD}</math> and call this intersection point <math>Y</math>. Note that triangle <math>BCD</math> is an isosceles triangle so angles <math>CDB</math> and <math>CBD</math> are each <math>5</math> degrees. Since <math>AB</math> equals <math>BC</math>, angle <math>BAC</math> had to equal <math>55</math> degrees, thus making angle <math>AYB</math> equal to <math>60</math> degrees. We can also find out that angle CYB equals <math>120</math> degrees. Extend point <math>C</math> such that it lies on the same level of segment <math>AB</math>. Call this point <math>E</math>. Since angle <math>BEC</math> plus angle <math>CYB</math> equals <math>180</math> degrees, quadrilateral <math>YCEB</math> is a cyclic quadrilateral. Next, draw a line from point <math>Y</math> to point <math>E</math>. Since angle <math>YBC</math> and angle <math>YEC</math> point to the same arc, angle <math>YEC</math> is equal to <math>5 degrees</math>. Since <math>EBD</math> is an isosceles triangle(based on angle properties) and <math>YAE</math> is also an isosceles triangle, we can find that <math>YAD</math> is also an isosceles triangle. Thus, each of the other angles is <math>\frac{180-120}{2}=30</math> degrees. Finally, we have angle <math>BAD</math> equals <math>30+55=\boxed{85}</math> degrees.
 
  
−
==Solution 6==
+
Now, by identity <math>\sin(\theta) = \sin(180^\circ-\theta)</math>, <math>\sin(65^\circ) = \sin(115^\circ)</math>
  
−
First, connect the diagonal <math>DB</math>, then, draw line <math>DE</math> such that it is congruent to <math>DC</math> and is parallel to <math>AB</math>. Because triangle <math>DCB</math> is isosceles and angle <math>DCB</math> is <math>170^\circ</math>, the angles <math>CDB</math> and <math>CBD</math> are both <math>\frac{180-170}{2} = 5^\circ</math>. Because angle <math>ABC</math> is <math>70^\circ</math>, we get angle <math>ABD</math> is <math>65^\circ</math>. Next, noticing parallel lines <math>AB</math> and <math>DE</math> and transversal <math>DB</math>, we see that angle <math>BDE</math> is also <math>65^\circ</math>, and subtracting off angle <math>CDB</math> gives that angle <math>EDC</math> is <math>60^\circ</math>.
+
Therefore, <math>\sin(115^\circ-x) = \sin(x-55^\circ).</math> This equation is only satisfied by option <math>\boxed{\text {(C) } 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.
 +
 
 +
~Raghu9372
 +
 
 +
==Solution 4 (Cheese)==
 +
 
 +
Using a protractor and rule, draw an accurate diagram ([https://i.imgur.com/YIafDgs.jpg/img Example Diagram]). <math>\angle BAD</math> looks slightly less than <math>90</math> degrees. Therefore the answer is <math>\boxed{\textbf{(C) } 85}</math> as <math>85</math> is slightly less than <math>90</math>.
 +
 
 +
 
 +
 
 +
~bobthegod78
 +
 
 +
~Reworded by [https://artofproblemsolving.com/wiki/index.php/User:South 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 <math>\theta</math>, and AB=BC=CD=1
 +
 
 +
The y value of from B-A is <math>\sin\theta</math>. The y value from C-B is <math>\theta-(180^\circ-70^\circ)=\theta-110^\circ</math>. The y value from D-C is <math>\theta-110^\circ-(180^\circ-170^\circ)=\theta-120^\circ</math>
 +
 
 +
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: <cmath>\sin\theta+\sin(\theta-110^\circ)+\sin(\theta-120^\circ)=0</cmath>
 +
 
 +
from here we expand out the trig expressions using sin addition and isolate <math>\theta</math>
 +
 
 +
<cmath>\sin\theta+\sin(\theta-110^\circ)+\sin(\theta-120^\circ)=0</cmath>
 +
<cmath>\sin\theta+\sin(\theta)\cos(110^\circ)-\cos(\theta)\sin(110^\circ)  +  \sin(\theta)\cos(120^\circ)-\cos(\theta)\sin(120^\circ)=0</cmath>
 +
 
 +
 
 +
 
 +
<cmath>1+\cos(110^\circ)-\cot(\theta)\sin(110^\circ)+\cos(120^\circ)-\cot(\theta)\sin(120^\circ)=0</cmath>
 +
<cmath>\cot(\theta)=\frac{1+\cos(110^\circ)+\cos(120^\circ)}{\sin(110^\circ)+\sin(120^\circ)}</cmath>
 +
 
 +
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).
 +
 
 +
<cmath>\cot(\theta)=\frac{1+\cos(120^\circ-10^\circ)+\cos(120^\circ)}{\sin(120^\circ-10^\circ)+\sin(120^\circ}</cmath>
 +
 
 +
<cmath>\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}</cmath>
 +
 
 +
<cmath>\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}}</cmath>
 +
 
 +
<cmath>\cot(\theta)=\frac{1-\cos(10^\circ)+\sqrt{3}\sin(10^\circ)}
 +
{\sqrt{3}\cos(10^\circ)+\sin(10^\circ)+\sqrt{3}}</cmath>
 +
 
 +
since the expression still isn't simplified, notice that using the double angle identity on cosine can be used to cancel the 1, and <math>\sqrt{3}</math>
 +
 
 +
Let: <math>\cos(5^\circ)=a</math>, <math>\sin(5^\circ)=b</math>
 +
 
 +
<cmath>\cot(\theta)=\frac{1-(1-2b^2)+\sqrt{3}\cdot(2ab)}
 +
{\sqrt{3}(2a^2-1)+2ab+\sqrt{3}}</cmath>
 +
 
 +
<cmath>\cot(\theta)=\frac{2b^2+2ab\sqrt{3}}
 +
{2a^2\sqrt{3}+2ab}</cmath>
 +
 
 +
<cmath>\cot(\theta)=\frac{2b(b+a\sqrt{3})}
 +
{2a(a\sqrt{3}+b)}</cmath>
 +
 
 +
<cmath>\cot(\theta)=\frac{b}{a}</cmath>
 +
 
 +
<cmath>\cot(\theta)=\tan(5^\circ)</cmath>
 +
 
 +
<cmath>\cot(\theta)=\cot(85^\circ)</cmath>
 +
 
 +
<cmath>\theta=85^\circ</cmath>
 +
~sahan
 +
 
 +
==Solution 6 (guess and check)==
 +
 
 +
Obtain: <math>\sin\theta+\sin(\theta-110^\circ)+\sin(\theta-120^\circ)=0</math> from solution 5
 +
We now guess <math>\theta=85^\circ</math> and try to verify
 +
 
 +
<cmath>\sin85^\circ+\sin-25^\circ+\sin-35^\circ=0</cmath>
 +
<cmath>\cos5^\circ=\sin25^\circ+\sin35^\circ</cmath>
 +
<cmath>\cos5^\circ=\sin(30^\circ-5^\circ)+\sin(30^\circ+5^\circ)</cmath>
 +
<cmath>\cos5^\circ=2\sin30^\circ\cos5^\circ</cmath>
 +
<cmath>\cos5^\circ=\cos5^\circ</cmath>
 +
 
 +
==Solution 7 (alternate way to bash out algebra + trig identities) ==
 +
 
 +
Use equation from Solution 5:
 +
<cmath>\sin\theta+\sin(\theta-110^\circ)+\sin(\theta-120^\circ)=0</cmath>
 +
<cmath>\sin\theta+\sin(\theta-115^\circ+5^\circ)+\sin(\theta-115^\circ-5^\circ)=0</cmath>
 +
<cmath>\sin\theta+2\sin(\theta-115^\circ)\cos(5^\circ)=0</cmath>
 +
<cmath>\sin\theta-2\sin(\theta+65^\circ)\cos(5^\circ)=0</cmath>
 +
<cmath>\sin\theta=2\sin(\theta+65^\circ)\cos(5^\circ)</cmath>
 +
 
 +
now guess <math>\theta=85</math> so that the cos(5) is dealt with (and then check it works)
 +
 
 +
If you refuse to guess work through the following algebra :D
  
−
Now, because we drew <math>ED = DC</math>, triangle <math>DEC</math> is equilateral. We can also conclude that <math>EC=DC=CB</math> meaning that triangle <math>ECB</math> is isosceles, and angles <math>CBE</math> and <math>CEB</math> are equal.
+
Let: <math>\phi=\theta+65^\circ</math>
  
−
Finally, we can set up our equation. Denote angle <math>BAD</math> as <math>x^\circ</math>. Then, because <math>ABED</math> is a parallelogram, the angle <math>DEB</math> is also <math>x^\circ</math>. Then, <math>CEB</math> is <math>(x-60)^\circ</math>. Again because <math>ABED</math> is a parallelogram, angle <math>ABE</math> is <math>(180-x)^\circ</math>. Subtracting angle <math>ABC</math> gives that angle <math>CBE</math> equals <math>(110-x)^\circ</math>. Because angle <math>CBE</math> equals angle <math>CEB</math>, we get <cmath>x-60=110-x</cmath>, solving into <math>x=\boxed{85^\circ}</math>.
+
<cmath>\sin(\phi-65^\circ)=2\sin\phi\cos(5^\circ)</cmath>
 +
<cmath>\sin\phi\cos(65^\circ)-\cos\phi\sin(65^\circ)=2\sin\phi\cos(5^\circ)</cmath>
 +
<cmath>\cos(65^\circ)-\cot(\phi)\sin(65^\circ)=2\cos(5^\circ)</cmath>
 +
<cmath>\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)</cmath>
 +
<cmath>\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)</cmath>
 +
<cmath>\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))</cmath>
 +
<cmath>\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))</cmath>
 +
<cmath>\cot(\phi)=-\sqrt3</cmath>
  
 +
<cmath>\phi=150^\circ</cmath>
  
−
<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 + dir(360-65-115-55-30));
 
−
label("$65^\circ$",B + dir(180-32.5));
 
−
label("$x^\circ$",A + dir(42.5));
 
−
label("$5^\circ$",D + dir(360-60-2.5));
 
−
label("$60^\circ$",D + dir(360-30));
 
−
label("$60^\circ$",E + dir(360-150));
 
−
label("$5^\circ$",B + 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.
+
<cmath>\theta=150^\circ-65^\circ=85^\circ</cmath>
  
−
~Someonenumber011
+
<cmath>\angle BAD = 85^\circ</cmath>
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=23|num-a=25}}
 
{{AMC10 box|year=2008|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:31, 28 August 2024

Problem

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.

~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 $\boxed{\text {(C) } 85^\circ}$

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.

~Raghu9372

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 $90$.


~bobthegod78

~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).

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

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

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

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

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

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

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

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

\[\cot(\theta)=\frac{b}{a}\]

\[\cot(\theta)=\tan(5^\circ)\]

\[\cot(\theta)=\cot(85^\circ)\]

\[\theta=85^\circ\] ~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: $\phi=\theta+65^\circ$

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

\[\phi=150^\circ\]


\[\theta=150^\circ-65^\circ=85^\circ\]

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

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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