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Difference between revisions of "1971 AHSME Problems/Problem 31"

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<cmath>16x^2+8x-224=0</cmath>
 
<cmath>16x^2+8x-224=0</cmath>
  
−
<cmath>2x^2+x-56=0</cmath>
+
<cmath>2x^2+x-28=0</cmath>
  
 
<cmath>x=\frac{-1 \pm \sqrt{1-(4)(2)(-28)}}{2(2)}=\frac{-1\pm \sqrt{225}}{4}=\frac{-1 \pm 15}{4}</cmath>
 
<cmath>x=\frac{-1 \pm \sqrt{1-(4)(2)(-28)}}{2(2)}=\frac{-1\pm \sqrt{225}}{4}=\frac{-1 \pm 15}{4}</cmath>
  
 
Since the negative solution is an extraneous solution, we see that our solution is <math>x=\frac{7}{2}\Rightarrow\boxed{A}</math>
 
Since the negative solution is an extraneous solution, we see that our solution is <math>x=\frac{7}{2}\Rightarrow\boxed{A}</math>

Revision as of 09:28, 18 August 2022

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.

With that train of thought, we see that

\[BD^2=AD^2-AB^2\]

\[BD^2=4^2-1^2\]

\[BD^2=15 \Rightarrow BD = \sqrt{15}\]

To let us formulate that rest of our lengths, let $CD=x$. Then, similar to the above,

\[AC^2=AD^2-CD^2\]

\[AC^2=4^2-x^2=16-x^2\]

\[AC=\sqrt{16-x^2}\]

Now that we have all of the side lengths and diagonal lengths in one variable, we use Ptolemy's Theorem to finish from here:

\[4\cdot1 + x\cdot1 = \sqrt{15}\sqrt{16-x^2}\]

\[x+4=\sqrt{240-15x^2}\]

\[x^2+8x+16=240-15x^2\]

\[16x^2+8x-224=0\]

\[2x^2+x-28=0\]

\[x=\frac{-1 \pm \sqrt{1-(4)(2)(-28)}}{2(2)}=\frac{-1\pm \sqrt{225}}{4}=\frac{-1 \pm 15}{4}\]

Since the negative solution is an extraneous solution, we see that our solution is $x=\frac{7}{2}\Rightarrow\boxed{A}$

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