Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | Quadrilateral <math>ABCD</math> has <math>AB = BC = CD</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 1== | |
− | + | * Note: This solution requires the use of cyclic quadrilateral properties but could be a bit time-consuming during the contest. | |
− | |||
− | |||
− | Note that | + | To start off, draw a diagram like in solution one 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. |
− | + | 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. | |
− | + | 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>. | ||
− | |||
<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); | ||
− | + | draw(E--C); | |
− | + | draw(B--D); | |
− | draw(B-- | + | 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, | + | label("$C$",C,SE); |
label("$D$",D,N); | label("$D$",D,N); | ||
− | label("$ | + | 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> | <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--B--C--D--cycle); | ||
− | + | draw(A--C); | |
− | draw( | + | 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); | |
− | label("$ | + | 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>. |
− | + | ||
+ | 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> | ||
+ | |||
+ | <math>\implies \frac{\sin(115^\circ-x)}{\sin(65^\circ)} = \frac{\sin(x-55^\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 <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 (best solution /j)== | ||
+ | |||
+ | Draw an accurate diagram and it looks like the angle is a bit less than 90, so answer is C. | ||
+ | [img]https://i.imgur.com/YIafDgs.jpg[/img] | ||
+ | |||
+ | ~bobthegod78 | ||
+ | ==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 | ||
+ | |||
+ | Let: <math>\phi=\theta+65^\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> | ||
+ | |||
+ | |||
+ | <cmath>\theta=150^\circ-65^\circ=85^\circ</cmath> | ||
− | = | + | <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 01:33, 24 November 2022
Contents
Problem
Quadrilateral has , and . What is the degree measure of ?
Solution 1
- 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 one and label the points. Create lines and . We can call their intersection point . Note that triangle is an isosceles triangle so angles and are each degrees. Since equals , angle equals degrees, thus making angle equal to degrees. We can also find out that angle equals degrees.
Extend and and let their intersection be . 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 degrees. 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.
~Minor edits by BakedPotato66
Solution 2
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
Solution 3(Using Trig.)
Let the unknown be .
First, we draw diagonal and . is the intersection of the two diagonals. The diagonals each form two isosceles triangles, and .
Using this, we find: and . Expanding on this, we can fill in a couple more angles. , , , .
We can rewrite and in terms of . and .
Let us relabel and .
By Rule of Sines on and respectively, , and
In a more convenient form,
and
Now, by identity ,
Therefore, This equation is only satisfied by option
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 (best solution /j)
Draw an accurate diagram and it looks like the angle is a bit less than 90, so answer is C. [img]https://i.imgur.com/YIafDgs.jpg[/img]
~bobthegod78
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 , and AB=BC=CD=1
The y value of from B-A is . The y value from C-B is . The y value from D-C is
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:
from here we expand out the trig expressions using sin addition and isolate
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).
since the expression still isn't simplified, notice that using the double angle identity on cosine can be used to cancel the 1, and
Let: ,
~sahan
Solution 6 (guess and check)
Obtain: from solution 5 We now guess and try to verify
Solution 7 (alternate way to bash out algebra + trig identities)
Use equation from Solution 5:
now guess 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:
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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