# CTK Insights

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19 Jun

### Climbing Pyramidal Slopes

On a pyramid having a square base and edge length $a$ path ascends from one of the bottom vertices to the apex at a slope of $25\%.$ Determine the height $h$ of the pyramid, the length $c$ of the first section of the path, the lengths $b$ and $a'$ as well as the total length $L$ of the path, all of them in terms of $a.$

(This problem, as I was informed by Peter Avxbybcbhybf (Petros Avxbybcbhybf), has been used at "the preparatory math camp for prospective IMO applicants to go to Korea for the IMO" (2000).)

It is not very steep and may be even tedious, but - at the end - the answer (summit) proves to be somewhat simpler than the climb that led there.

I slipped once and got an answer that included the golden ratio. Since the latter commonly pops up in unexpected situations, I was not at all surprised. However, I noticed in time that the golden ratio would lead the path downhill.

Tumbling into the golden ratio had its upside: if it were not for that, I would probably give up and return home without reaching the summit. As it happened, correcting the mistake took me all the way up there.

Solution

The easiest part of the solution is finding the height of the pyramid. This is the vertical leg of the right triangle with the base $\displaystyle\frac{a}{2}$ and hypotenuse $\displaystyle\frac{\sqrt{3}}{2}.$

What it comes out to is $\displaystyle h=\frac{\sqrt{2}}{2}a.$ For the rest of the solution I'll depend on the following diagram:

One thing known right away is that $\displaystyle u=\frac{c}{4}.$ From this and the Pythagorean theorem we can find the bases $v$ and $w$ in terms of $c$, and then express the latter as a function of $a.$

$\displaystyle v = \frac{\sqrt{15}}{4}c,$

$\displaystyle w = u\mbox{cot}(\angle EBD) =\frac{c}{4}\frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{12}c,\space \mbox{because}\space\angle EBD=60^{\circ},$

$\displaystyle a = v+w = \frac{3\sqrt{15}+\sqrt{3}}{12}c,$

$\displaystyle c = \frac{12a}{3\sqrt{15}+\sqrt{3}} = \frac{3\sqrt{15}-\sqrt{3}}{11}a,$

$\displaystyle w = \frac{\sqrt{3}}{12}\frac{3\sqrt{15}-\sqrt{3}}{11}a = \frac{3\sqrt{5}-1}{44}a,$

$\displaystyle b = 2w = \frac{3\sqrt{5}-1}{22}a,$

$\displaystyle a' = a - b = \frac{23-3\sqrt{5}}{22}a,$

$\displaystyle q = \frac{a'}{a}=\frac{23-3\sqrt{5}}{22},$

$\displaystyle \frac{1}{1-q} = \frac{3\sqrt{5}+1}{2},$

$\displaystyle L = \frac{c}{1-q} = \frac{(3\sqrt{15}-\sqrt{3})(3\sqrt{5}+1)}{22}a=2\sqrt{3}a.$