3 Problems From China With Outside The Box Thinking
Note to long-time readers: Puzzle 1 is a new problem. Puzzles 2 and 3 are re-posts added to this compilation. (YouTube is preferring longer videos, and it is a good time to re-share overlooked problems.)
Puzzle 1
This is adapted from a 6th grade cram school problem in China. A solid figure is made by stacking several cubical blocks with a side length of 1 cm. The views from the front, the side, and above are shown. What is the volume of the solid?
Puzzle 2
A square is inscribed along the right angle of a right triangle, as shown, with vertical and horizontal lengths to the corners of 5 and 20. What is the area of the square?
Puzzle 3
Thanks to Eric from Miami for suggesting this problem and sending a solution!
In the quadrilateral ABCD, angle A = 90°, angle ABD = 40°, angle BDC = 5°, angle C = 45°, and the length of AB is 6. Find the area of the quadrilateral ABCD.
As usual, watch the video for solutions.
3 Problems From China With Outside The Box Thinking
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Answer To 3 Problems From China With Outside The Box Thinking
(Pretty much all posts are transcribed quickly after I make the videos for them-please let me know if there are any typos/errors and I will correct them, thanks).
Puzzle 1
Think about a 3x3x3 cubical array of 27 cubes. From the front view, we must eliminate the 3 cubes on the top right row. From the side view, we remove 2 rows of 2 cubes, a total of 4 cubes, from the left and right top rows. From the top view we must remove all corner cubes, which is 4 groups of 2 cubes, or 8 cubes in total.
Thus there are 27 - 3 - 4 - 8 = 12 cubes in the figure. Each cube has a volume of 1 cm3 so the volume of the shape is 12 cm3.
Reference
Post on X
https://x.com/AonghasCrowe/status/1843790315314130987
Puzzle 2
A textbook solution involves similar triangles. Suppose the square has a side length equal to x. Then the green triangle is similar to the white triangle.
So we have:
x/20 = 5/x
x2 = 20(5) = 100
But x2 is exactly the area of a square with side length x, so this is the answer!
Another interesting method uses outside the box thinking. Construct a rectangle containing the original large triangle, and extend the horizontal and vertical lines from the square to the rectangle.
The two large triangles are equal as each is half of the large rectangle. But we also see the two triangles marked 2 in the upper left are congruent, as as the two in the lower right marked 3. Subtracting equal areas from both large triangles means the remaining areas are equal. So the square’s area must exactly be that of the rectangle. But what is the rectangle’s area? Its dimensions are already specified from the given information! It has a width of 20 and a height of 5, so its area is 20(5) = 100. Therefore the square has an area equal to 5, and that’s the answer!
Puzzle 3
I would have solved the problem using trigonometry, which is beyond what a 5th grader would have learned. Still, I will present this complicated solution to show it is possible. Then I will reveal the neat trick which is how 5th graders were expected to solve it.
First we will calculate some angles and lengths.
angle ADB = 90° - 40° = 50°
angle DBC = 180° - 45°- 5° = 130°
AD = 6 tan 40°
BD = 6/cos 40°
Then we can use the law of sines to calculate CB:
CB/BD = (sin 5°)/(sin 45°)
CB = (6/cos 40°)(sin 5°)/(sin 45°)
We can then calculate the areas of triangles ABD and CBD. Triangle ABD is a right triangle, so its area is half its base times its height. Triangle CBD can be solved using the formula (side 1)(side 2)(angle between)/2. So we get:
area ABD = 6(6 tan 40°)/2
area CBD = (6/cos 40°)[(sin 5°)/(sin 45°)](sin 130°)(1/2)
The total area is the sum of these, which can be found numerically on a calculator. Remarkably, the answer simplifies tremendously-it is an integer!
area ABCD = 18
So now the question is, how would 5th graders have found the answer? Here’s the incredible method.
The trick: flip triangle CBD
If we flip triangle CBD, side BD remains in the same position, but we flip the other sides. Here is the shape we end up creating:
Now we can calculate some angles in this new figure:
angle ADC = 50° + 130° = 180°
angle ABC = 40° + 5° = 45°
angle ACB = 45°
Notice this is a right triangle with two equal 45 degree angles-so this is an isosceles right triangle!
The area of this shape equals quadrilateral ABCD (as we have only re-arranged the areas). So the other leg of this triangle AC also has a length of 6, and its area will be:
6 × 6 / 2 = 18
Amazing! This is unlike any math problem I’ve seen in class, a competition, or on a test. It’s quite a neat trick!
Thanks to Nestor Abad!
He figured out how the trigonometric expression simplifies to 18. Perhaps this is a good exercise for trigonometry students:
https://drive.google.com/file/d/1Ll1SfK94N9mgqPRzxaNGGyv9u-YTAHVJ/view?usp=sharing
https://mindyourdecisions.com/blog/2025/01/28/3-problems-from-china-with-outside-the-box-thinking/
https://mindyourdecisions.com/blog/?p=37545