Galliv-Ant up the Pyramid

By the Pythagorean Theorem,

$a=\sqrt{2}$ .

We then 'flatten' the pyramid. Now visualize $\triangle ABE$ and $\triangle BEC$ . $\cos \angle AEB=\cos \angle CEB$ . (As they are congruent.)

We know that $AB=2$ and $AE=EB=\sqrt{3}$ .

By the Law of Cosines,

$-6\cos \angle AEB=-2$ ,

$\cos \angle AEB=\cos \angle CEB=\dfrac{1}{3}$ .

And $\cos \angle CEA= (\dfrac{1}{3})^{2}-1=-\dfrac{7}{9}$ .

Then again by the Law of Cosines,

$(AP)^{2}=(\sqrt{3})^{2}+(\dfrac{\sqrt{3}}{2})^{2}-3(-\dfrac{7}{9})=\dfrac{73}{12}$ .

And $73+12=\boxed{85}$ .

Unfolded image of EABC

• Unfold the pyramid so you can see, in flat, the shape $EABC$ .
• Let $M$ be the midpoint of $EC$ .
• Draw a straight line from point $A$ to $M$ .
• Let $h$ = length of $AM$ . The value of $h^2$ is our target.
• Let $s$ = length of $EA$ . Then $s = \sqrt{1^{2} +1^{2} +1^{2}} = \sqrt{3}$
• Let $t$ = length of $EM$ . Since $EA = EC$ , and $EM = {EC \over 2}$ , then $t = {EC \over 2} = {\sqrt{3} \over 2}$ .
• Let $\alpha = \angle AEM$ .

Then, using cosine law: $h^2 = s^2 + t^2 - 2 s t \cos(\alpha)$

To get $\cos(\alpha)$ :

• Label the midpoint of $BC$ as $N$ and draw the line $EN$ . Its length is $EN = \sqrt{1^2 + 1^2} = \sqrt{2}$
• Let $\beta = \angle NEC$ . Notice that $\alpha = 4\beta$ .
• Then we get $\cos(\beta) = \frac{adjacent}{hypotenuse} = \frac{EN}{EC} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{2 \over 3}$
• Thus: $\begin{aligned} \cos(\alpha) &= \cos(4\beta) \\ &= 2\cos^2(2\beta) - 1 \\ &= 2\left(2 \cos^2(\beta) - 1\right)^2 - 1 \\ &= 2\left(2\left( \sqrt{2 \over 3} \right)^2 - 1\right)^2 - 1 \\ &= 2\left({4 \over 3} - 1\right)^2 - 1 \\ &= {2 \over 9} - 1 \\ \cos(\alpha) &= -{7 \over 9} \\ \end{aligned}$

Back to our 1st formula: $\begin{aligned} h^2 &= s^2 + t^2 - 2st \cos(\alpha) \\ &= \sqrt{3}^2 +\left({\sqrt{3} \over 2 }\right)^2 - 2\sqrt{3}\left({\sqrt{3} \over 2 }\right) \left(-{7 \over 9}\right) \\ &= 3 + \frac{3}{4} + \frac{21}{9} \\ &= 3 + \frac{3}{4} + \frac{7}{3} \\ &= \frac{36 + 9 + 28}{12} \\ h^2 &= \frac{73}{12} \\ \end{aligned}$

Therefore, $73 + 12 = \boxed{85}$

To solve this problem imagine that the pyramid having square base ABCD is made of cardboard which can be unfolded. Let the apex of the pyramid be E. Cut the pyramid open along the line AE. Unfold it and place it flat. Then, ABCE forms a quadrilateral where AB = BC = 2, AE = BE = CE = √3. P is the mid-point of CE, hence PE = √3/2. The shortest path is therefore the straight line AP on the flat quadrilateral. Using the cosine formula, Cos ∠AEB = 1/3. Hence cos ∠AEC = (1/3)^2 – 1 = - 7/9. Then using cosine formula in ΔAEP, AP^2 = 73/12. Answer = 85.