3 outside the box problems from China

How many can you solve?

0:00 puzzle 1
3:00 puzzle 2
7:07 puzzle 3

References
Thanks to Eric from Miami for puzzle 3
Trig verification in puzzle 3

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[essemque]

Oh that triangle flip on #3 is clever!!

[Omsip]

Guys, look up the word “stacking” before commenting there could be holes inside!!1

[roginutah]

I like the out-of-the-box solution on #2. Good job on #3, too.

[kavyagupta_]

The last solution of the puzzle 3 got me 💀 truly a genius soln. No required complications.

[LLlap]

You can also do it backwards which seems easier. -12 on the above, -4 on the side and -1 on the front.

[WillRennar]

How I solved the 2nd puzzle:

(x+5)(x+20)/2 = (x^2 + 25x + 100)/2 = area of triangle

5x/2 = 2.5x = area of upper triangle
20x/2 = 10x = area of lower triangle
x^2 = area of square
2.5x + 10x + x^2 = x^2 + 12.5x = area of all 3 shapes combined = area of triangle

12.5x + x^2 = (x^2 + 25x + 100)/2
multiply both sides by 2
2x^2 + 25x = x^2 + 25x + 100
subtract (x^2 + 25x) from both sides
x^2 = 100 = area of square

[wayneyadams]

My favorite solutions are the outside the box solutions. The last problem was the best. Grinding it out was treeibly complex and time consuming, but the second method was ridiculously easy.

[matthewmichaelson]

Lol that last one was a trip. 20 trig calculations or flip the flipping triangle!

[heqitao]

What a great set of puzzles. I love the simple solutions in #2 and #3.
Thank you for sharing.

[Hd]

12:00 This triangle here would e called ACB because you need to name points counterclockwise.
The cross multiplication concept is pretty bad for understanding math. It’s just simpler to understand if you explain that both sides of an equation stay equal if you perform an operation on both sides.

[shrijagadish]

Puzzle 3 can be generalized. For any angle b (that is ABD) less than 45 deg, angle BDC should be 45 - b and angle BCD should be 45 degrees.
In the given problem b = 40 deg, so BDC = 45 - 40 = 5 deg.
Another example b = 30 deg. So BDC = 45 - 30 = 15 deg.
For all these cases the triangle flip can be done to get an isosceles right triangle with the same area.

[ryokubudo]

Puzzle no 1 was easy, puzzle no 2 especially second method thinking out of the box blow my mind! Thank you very much for this joyful puzzle!

[iamadooddood]

For puzzle 1, you can remove the middle middle cube and/or the bottom middle cube and still get the same three views. These two cubes are obscured from view so their presence or absence doesn't matter.

[Vienticus]

2:00 most interesting way to solve Rubik's Cube ever.

[cubefromblender]

3:24
Lemme try
Since those triangles are similar, and one of the sides of each of them is also a side of the square
5/x=x/20 (×20)
100/x=x (×x)
100=x^2
Thus is already the area but the side length is √100=10
The side length isn't -10 because it's geometry

[tedwatson]

In the first puzzle, why are you assuming the unshown center bottom 2 spaces are filled with cubes? Unless they tell us that we cannot definitively say that they are there.

[_Loki__Odinson_]

What I don't understand is how on earth is it possible for a 10 year old to solve that last question

[razaele]

The first one is just descriptive geometry, I don't really get why it's it viral 🤔

[mathmannix]

For those who are saying the first puzzle can have only 10 or 11, I would have thought that, except that the instructions specified it was SOLID, i.e. not hollow.

[dannypipewrench]

Be careful! That first problem might summon a technical drawing professor. Do you want to be haunted by AutoCAD?