Solving a 3d geometry problem for PhD students

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[jeffgerken]

I thought about this problem for quite a while, and came up with a simpler method for determining the value of t. Assume that there are four hexagons around the square. When all of them are bent up by the angle t, they will form a square above the initial square. If the initial square has sides of 2, as in your example, the new square will have sides of 4, so that the "shadow" of that square will extend beyond the original square by 1 unit. A triangle from the center of one side of the square to the center of the corresponding hexagon will have a base of 1 unit and a hypotenuse of sqrt(3) units, so the cosine of the angle between the plane of the square and the hypotenuse of the triangle will be 1/sqrt(3), and the angle will be ~54.74 degrees.

[Bashamo]

Using only the test paper, some scissors, tape, and a protractor, the answer was determined experimentally.

[banikrai]

Thanks Presh on making a video on this question. I instantly sat down to watch this video as soon as I got your mail. It was quite helpful and worth the long wait.

[ОлександрМихальчук-ьо]

Wow, that was really cool. Especially 15 years after leaving school - that was fantastic

[jansimacek]

I tried solving it considering the fact that this is a part of a truncated octahedron:
1. the side of the hexagon is 1/3
2. by "completing" the hexagon into a triangle we get an equilateral triangle who's side has the length 1
3. this triangle is a wall of the untruncated version of the truncated octahedron
4. consider one of the octahedrons 2 constituent pyramids - all its edges have the length 1 and it's diagonal has the length sqrt(2)
5. using the pythagorean theorem, the pyramid's height is sqrt(1-1/2) = sqrt(2)/2
6. the line segment XY, where X is the center of the pyramid's base and Y is the center of one of the base's sides, has the length 1/2
7. the angle at XYZ, where Z is the peak of the pyramid, equal to 54°44'8.2"
8. the plane of the square at the beginning is parallel to the plane of the pyramids base, therefore XYZ must be the angle were looking for

The first time I tried this method I put in the wrong values at step five (I was octahedron's considering the side length being 3, so there was more room for error). Then tried another solution, which involved finding the ratio which determines how moch the hexagon's shadow scales as you turn it upwards, which gave me the correct solutions.
Guess we're all getting PhD's

[WestExplainsBest]

I get the feeling Presh is a big fan of geometry problems. That makes at least 2 of us.

[.]

If you look at one of the hexagons sides tilting up, the question becomes, at what angle does a 30º angle look like a 45º angle? From this, you can draw a tetrahedron of a 30-60-90 triangle being projected onto a plane to make a 45-45-90 triangle. Filling in the side length ratios gives you arccos(1/sqrt3)

[nightbot.]

I cracked this problem too. I tracked the locus that the two corners traced and without loss of generality labelled the side length 1. I deduced that they'd meet on a point above the line angled at 45° to the base. I used cosine rule and obtained cos(t)=1/sqrt3. And tis approximately 54.74°.
Thanks Presh for the wonderful and thought-provoking problem
From Ireland 🇮🇪

[Jtretta]

I solved this by solving for the intersection between the circles swept out by E and F while folding. Solving for z then setting the circle equations equal and doing basic canceling nets you the sqrt(2) z height. From there the sqrt(2)/2 ratio practically screams 45°.

[Anmol_Sinha]

Imagining 3 dimensions is hard. Let alone solving them 😅
Respect to your editing which managed to present the solution in an understandable way.

[twwc]

I treated the various sides as vectors in 3-d space and used quaternion methods for the rotations, but in the end it works out to basically the same solution technique as yours. Your method is probably simpler, but I'm glad I was finally able to use quaternions to solve a Mind Your Decisions problem!

[peterzerfass]

45° seems obvious because you can extend the object with multiple copies of itself due to symmetry considerations into a closed shape with squares facing in the +x/-x, +y/-y and +z/-z direction...which forces AE to have a 45° angle with the plane of the square.

[VinTheFox]

I was able to solve it without going into a 3d coordinate system - I just worked in whatever plane I needed to work in. For part 2 you can just focus on the plane of ACE (=ACF). Immediately you have the right triangle with hyp=2 and height rt(2).

[SuperYoonHo]

Thanks so much Presh Talwalker for sharing with us this wonderful video! You are making us real smart!

[maxryder]

My first thought seems like a simple solution. Simply give the “y coordinate” of E as a function of the angle, and solve for when this is equal to the “x coordinate”, which is a constant. By symmetry between E and F, this is the solution.

[Peter-osvl]

Since the hexagon consists of 6 equilateral triangles, you can see with a little spatial imagination that the angles you are looking for correspond to the interior angles of a pyramid with a square base and equal side lengths. The angles are therefore arccos(1/sqrt(2)) and arccos(1/sqrt(3)).

[sagarvekariya]

One of the sol can be

U take parallopiped then try for volume of it by doing box product of it ..and then equat it with the area×asint...where area=(a)^2.... that it you get it t= 45

[schungx]

Hexagon + square is one regular solid, so it has rotational symmetry. Square pieces are paired front-to-back, and another one at 90 degrees offside. As each square can serve as a base reference, by the principle of symmetry, each diagonal line segment must connect at 45 degrees in order to preserve the symmetry. So the angle AE makes must be 45 degrees and cannot be anything else.

[fabrizioruizdiaz]

I found the answer by using the power of vectors, as the hexagons have 120° in FAD and EAB then, those will remain constant with the rotation, so, using cos directors I found the angle with Z axis, so the answer to question B is 90 - the number (45°), now for question a I used cross multiplication with the vector I found and AD, then the same analisis with the same z axis. Had a lot of fun with this one.

[justanotherrandomguy......]

Well i have a better solution
We can solve it through 3d geometry considering 3 lines from A. Angle FAD and EAB are 120° and let 3rd be alpha
Now by using(line makes angle α, β, γ with the axes then cos²α+cos²β+cos²γ=1
Just substitute and get 3rd angle =45
So simple