Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.3 National Olympiad, round 2 Prove it Romania

Let $ABCDEF$ be a regular hexagon with side length $a$ . At point $A$ , the perpendicular $AS$ , with length $2a\sqrt{3}$ , is erected on the hexagon's plane. The points $M, N, P, Q$ , and $R$ are the projections of point $A$ onto the lines $SB, SC, SD, SE$ , and $SF$ , respectively.

a) Prove that the points $M, N, P, Q, R$ lie in the same plane.

b) Find the measure of the angle between the planes $(MNP)$ and $(ABC)$ .

Solution

a) Using the Three Perpendiculars Theorem, from $SA \perp (ABC)$ and $AB \perp BD$ , it results $SB \perp BD$ . Since $BD \perp AB$ and $BD \perp SB$ , it follows that $BD \perp (SAB)$ , hence $BD \perp AM$ .

Since $AM \perp SB$ , it results $AM \perp (SBD)$ , hence $AM \perp SD$ . We also have $SD \perp AP$ , therefore we obtain that $SD \perp (AMP)$ . In a similar way one can show that $SD \perp (ARP)$ , $SD \perp (ANP)$ and $SD \perp (AQP)$ , therefore the points $M, N, P, Q, R$ lie in the same plane.

b) Because $MR \parallel BF$ , it follows that the intersection between the planes $(MNP)$ and $(ABC)$ is the line $d$ parallel to $BF$ and passing through $A$ . Since $d \perp SA$ and $d \perp AD$ , it results that $d \perp (SAD)$ , hence $d \perp AP$ . Therefore the angle between the planes $(MNP)$ and $(ABC)$ equals $\widehat{PAD}$ .

Using the Pythagorean Theorem, one obtains $SD = 4a$ , hence $m(\widehat{PDA}) = 60^\circ$ and $m(\widehat{PAD}) = 30^\circ$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.