Twisting a Prism

Geometry Level 5

A certain kind of prism has one square face (shown on the bottom) and one equilateral triangle face (shown on top), both with an area of 1, parallel to each other and sharing a common axis (red) perpendicular to both and passing through the centers of both. The faces are separated by a distance of 1. The edges connect the vertices of both the square and the equilateral triangle, as shown in the graphic. One of the altitudes of the equilateral triangle is parallel with one of the diagonals of the square. The volume of this prism is greater than 1.

A typical cross section of this prism parallel to the faces is an irregular heptagon, as shown in dark blue, this one being exactly halfway between the ends.

If the equilateral triangle face is rotated about the red axis by a certain angle (either way) while the square face remains stationary (and both faces remain parallel), the volume of the prism is reduced. When this volume becomes exactly 1, the same as a unit cube, this angle of twist can be exactly expressed as follows:

$\theta =\arcsin\left( \sqrt { a } -b \right) ,$ where $a$ and $b$ are positive integers with $\sqrt a> b$ . Find $a+b$ .

Note: Assume that the blue edges stretch or shrink as the prism is twisted, so that all the faces of the prism stay flat.

The answer is 4.

1 solution

Michael Mendrin
Mar 14, 2017

This is what it looks like when this prism is twisted until its volume has become exactly $1$ . The angle of twist is exactly

$\theta =ArcSin\left( \sqrt { 3 } -1 \right)$

so that the answer is $4$

Because this is a prism with a ruled surface connecting its two faces, we only need to find the angle of twist that will produce a cross-section area of exactly $1$ halfway between the faces, as indicated by the irregular heptagon in cyan. There is no need to compute any volumes or any other cross section area. The prismoidal formula is as follows

$V=\dfrac { 1 }{ 6 } h\left( { A }_{ 1 }+4{ A }_{ 2 }+{ A }_{ 3 } \right)$

where $h$ is the altitude, ${ A }_{ 1 }, { A }_{ 3 }$ are the areas of the end faces while ${A}_{2}$ is the area of the cross-section halfway between the two end faces.

Edit: Here's an direct means of computing this angle, see the following graphic

First, we translate the equilateral triangle so that the orthographic view of both the equilateral triangle and the square shows them sharing a common vertex, $O$ . The volume will be the same because all of the cross-sections will be the same regardless of how the equilateral triangle is translated. The angle we seek is $\theta$ . At this angle, the cross-section (red) very near to the square has an approximately thin rectangular sliver subtracted from the right side of the square and another approximately thin rectangular sliver added to the top side of the square. If the difference between the widths of both approach $0$ , then the area of this cross section approaches $1$ . Since the sides of this prism is a ruled surface, if the area of the cross section anywhere not at the ends is $1$ , then all cross sections have an area of $1$ . The condition necessary for such difference between the widths to approach $0$ (or, more precisely, the ratio of the two approaches $1$ ) is that the vertical distance between points $A, B$ and the horizontal difference between points $C, D$ be equal. Thus, we have the equation to solve for $\theta$

$\;$
$sSin\left( \theta +75 \right) -1=1-sCos\left( \theta +15 \right)$
$\;$

where $s$ is the side of the equilateral triangle

$s=\dfrac { 2 }{ \sqrt [ 4 ]{ 3 } }$

This angle $\theta$ then works out to

$\theta =ArcSin\left( \sqrt { 3 } -1 \right)$

At this angle of twist, then all cross-sections of this prism have an area of $1$ which can be confirmed by tedious computations.

@Michael Mendrin - Thanks for all the awesome problems!! How do you come up with so many cool problems? Do you read a lot of math? Thanks!!

Christopher Criscitiello - 3 years, 9 months ago

Twisting a Prism

The answer is 4.

1 solution

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