1971 AHSME Problems/Problem 31

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]$

Quadrilateral $ABCD$ is inscribed in a circle with side $AD$ , a diameter of length $4$ . If sides $AB$ and $BC$ each have length $1$ , then side $CD$ has length

$\textbf{(A) }\frac{7}{2}\qquad \textbf{(B) }\frac{5\sqrt{2}}{2}\qquad \textbf{(C) }\sqrt{11}\qquad \textbf{(D) }\sqrt{13}\qquad \textbf{(E) }2\sqrt{3}$

Solution

Solution 1

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 $AOB$ . The cosine of that angle is 7/8. We use the double cosine angle to find angle $AOC$ :

$cos(2x) = 2 cos^2 (x)- 1$

Therefore, $cos(2x) = 17/32$

Applying the Law of Cosines to the triangle $DOC$ , we get

$CD^2 = 2^2 + 2^2 - 2(4)(-17/32)$ (because $cos(x) = -cos(180-x)$ )

Therefore, $CD^2 = 8 + 17/4$

and we find that is equal to $7/2$ , or option A!

Solution 2

By the properties of a circle, we see that angle $\text{DCA}$ and angle $\text{DBA}$ are both right angles. Thus, if we formulate the diagonal lengths of the cyclic quadrilateral using the Pythagorean theorem, we can finish the problem with Ptolemy's Theorem.