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 (Cyclic Quadrilateral)== | |
− | + | * 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 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. |
− | + | 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== | |
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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>. | 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>. | ||
Line 100: | Line 36: | ||
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("$E$",E,NE); | label("$E$",E,NE); | ||
Line 115: | Line 51: | ||
~Someonenumber011 | ~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 <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 (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 | ||
+ | |||
+ | 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 18:31, 28 August 2024
Contents
Problem
Quadrilateral has
,
and
. What is the degree measure of
?
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 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 (Cheese)
Using a protractor and rule, draw an accurate diagram (Example Diagram). looks slightly less than
degrees. Therefore the answer is
as
is slightly less than
.
~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 , 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.