2008 AMC 10B Problems/Problem 24
Quadrilateral has , and . What is the degree measure of ?
- 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
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.
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,
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.
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]
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
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
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