Five Geometry Puzzles That Will Test Your Problem-Solving Skills
Triangle and Five Rectangles
Here’s a simple problem to start off, posted to the subreddit r/askmath. Rectangle ABCD is made of five congruent rectangles arranged together, where the length of AD is 10. A point E is placed on CD. What is the area of triangle ABE?
Since each rectangle is congruent and two of the short sides make up a long side, we know that each rectangle must have a 2:1 aspect ratio. Calling the short side x and the long side 2x, we know that 2x + x + 2x = 10, or 5x = 10, so x = 2. The length of AB is 2 times the short side of 2, giving us 4. Assuming AB is the base, the height of the triangle is the width of the rectangle, which is 10.
Now we just apply the formula for the area of a triangle: A = (1/2) × base × height = (1/2)(4)(10) = 20. So the area of triangle ABE is 20.
Wood Piece Puzzle
This problem comes from an anonymous source. Suppose you have a piece of wood 12 units wide and 9 units tall, which is a filled rectangle except for a missing hole in the center. The hole is a rectangle 8 units wide and 1 unit tall. Your task is to cut up this piece of wood using certain rules and rearrange it to fit in a square region with side lengths of 10 units. You may use only two cuts on the piece of wood, where a cut must be a path of any shape that starts and ends on an edge of the piece of wood (a boundary between the wood and empty space) and otherwise stays entirely within the piece of wood without touching an edge.
To describe the solution, let’s use Cartesian coordinates. We’ll put the bottom-left corner of the piece of wood at the origin (0, 0). Starting at the point (2, 9), cut down one unit, right two units, and repeat these two motions to form a staircase pattern until you finish the cut at (8, 5). Next, starting at the point (4, 4), cut with the same repeated motion. This time, the cut will finish at the point (10, 0).
The piece of wood is now split in two. Take the right piece and translate (shift) it up one unit and left two units, fitting perfectly with the left piece. This creates a wood piece formation that slots exactly into the square region, completing the puzzle.
Inscribed Triangle Problem
This problem comes from the National Museum of Mathematics. Consider a square ABCD. A point E is drawn on line segment CD and a point F is drawn on line segment AD, forming triangle ABF, triangle BCE, and triangle DEF. The areas of these triangles are 4, 3, and 5 respectively. What is the area of triangle BEF?
We can start by trying to find the area of the whole square. Let’s call the square’s side length a. Consider triangle ABF, which has a height of a and a base length that we’ll call b. Applying the area formula, the area 4 = (1/2)(b)(a). We can solve for b by multiplying both sides by 2 and dividing by a, giving b = 8/a. So line segment AF has a length of 8/a. Since line segment AD must have a length of a (the square’s side length), line segment DF has a length of a − 8/a.
For triangle BCE, we can apply similar logic. The area 3 = (1/2)(a)(h), where h is the height. So h = 6/a. Thus segment CE has a length of 6/a, and segment DE has a length of a − 6/a.
Finally, knowing DE and DF, we can relate the area of triangle DEF to its side lengths: 5 = (1/2)(a − 8/a)(a − 6/a). Multiplying both sides by 2 and distributing on the right, then subtracting 10 from both sides and multiplying by a², we can substitute x = a² to get a quadratic equation. Since a² is the area of the square, we just have to solve for x.
Using the completing-the-square method, and noting that the three given triangle areas sum to 12 (so the total area of the square must be greater than 12), we find x = 12 + 4√6. So triangle BEF’s area equals the square’s area minus the three other triangles: (12 + 4√6) − 4 − 3 − 5 = 4√6, which is about 9.8.
Wertheimer’s Car Riddle
The following riddle comes from psychologist Max Wertheimer, who sent it to Albert Einstein in 1934. Imagine a car that ascends to the peak of a hill, traveling 1 mile at a speed of 15 mph. The car then descends from the other side of the hill, again traveling 1 mile, though not necessarily at the same speed. If the average speed of the car over the course of the entire trip was 30 mph, how fast did the car travel during the descending portion?
At first you might be inclined to assume that the answer is 45 mph, since the average of 15 and 45 is 30. But this is incorrect. Einstein himself was initially fooled by the riddle.
To find the true answer, we must look to the actual definition of average speed. We take the distance traveled and divide it by the time taken. So if we let the time taken for the whole trip be t, that gives us 30 mph = 2 miles / t.
Dividing both sides by 2 miles and taking the reciprocal, we find that the total trip time must have been 1/15 of an hour, or 4 minutes. Now we need to find the time taken for the first part of the trip. Using the same formula: 15 mph = 1 mile / t₁. Dividing both sides by 1 mile and taking the reciprocal, we get t₁ = 1/15 hour.
At this point you may have realized the twist. The car must have taken zero hours for the descending portion of the trip. Thus, if the problem has any answer at all, you could say the car traveled infinitely fast during the descending portion, basically teleporting to the bottom instantly.
If you’re unhappy with Wertheimer’s version of this problem and want a less absurd version, try seeing what would happen if the average speed over the entire trip were 20 mph, not 30. This is left as an exercise to the reader.
Shaded Area Problem
This problem is from the 2020 AMC 10B, Problem 14. Draw a regular hexagon with side length 2. Now for each side, draw a semicircle inside the hexagon with the vertices of the side as the semicircle’s endpoints. What is the area of the region bounded by the semicircles at the center of the hexagon?
Draw line segments between opposite vertices of the hexagon, and draw line segments between the midpoints of sides that are two sides apart, creating a tiling pattern of equilateral triangles. To find the area of a tile, we use the formula for the area of an equilateral triangle of side length s: A = (√3/4)s². This side length is one half the side length of the hexagon, so the area of one of these triangles is (√3/4)(1²) = √3/4.
Six of these triangles contain a piece of the blue region, so we just need to find the blue area contained in one of these triangles and multiply by six. Consider one of these partially blue triangles. It shares a side with one other triangle that touches a side of the hexagon. These two triangles form a diamond shape with total area of 2 × √3/4 = √3/2.
Now to find the blue area, we need to subtract off the area of the remaining region, known as a circular sector. The formula for the area of a circular sector is A = (1/2)θr², where r is the radius and θ is the central angle in radians. A full 360° revolution is 2π radians.
The radius of our circular sector is 1 and the angle is 1/6 of a full turn, which is 2π/6 = π/3 radians. Substituting yields A = (1/2)(π/3)(1²) = π/6.
So the blue area within our diamond shape is √3/2 − π/6. Multiplying by 6 to get the full blue area, we get our final answer: 3√3 − π, which is about 2.05.
(The original AMC answer choice stated this as 3√3 − π, using π rather than τ, but of course it’s the same value.)
Further Reading
- AMC 10B 2020 Problems on Art of Problem Solving
- National Museum of Mathematics (MoMath)
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