1971 AHSME Problems/Problem 31
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Problem 31
![[asy] size(2.5inch); pair A = (-2,0), B = 2dir(150), D = (2,0), C; draw(A..(0,2)..D--cycle); C = intersectionpoint(A..(0,2)..D,Circle(B,arclength(A--B))); draw(A--B--C--D--cycle); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,N); label("$D$",D,E); label("$4$",A--D,S); label("$1$",A--B,E); label("$1$",B--C,SE); //Credit to chezbgone2 for the diagram[/asy]](http://latex.artofproblemsolving.com/2/3/f/23f9d4352df55f722a212cc274b8942934ad6fb4.png)
Quadrilateral 
is inscribed in a circle with side 
, a diameter of length 
. If sides 
and 
each have length 
,
then side 
has length
Solution
Note that the length 4 forms a semicircle. We can then use the Law of Cosines. Take the center and form a line segment with the other two points.
Let's find the cosine of angle 
. The cosine of that angle is 7/8. We use the double cosine angle to find angle 
:
Therefore, 
Applying the Law of Cosines to the triangle 
, we get
(because 
)
Therefore,
and we find that is equal to 
, or option A!
