Pythagorean Theorem Challenge Problems

Unfold the two faces the ant crosses so they lie flat. In the flat picture, the shortest route is a straight line. Try each way of pairing faces.

The ant must cross two faces that share an edge. Unfold them into a rectangle, and the shortest route is that rectangle's diagonal. There are three ways to pair the faces, depending on which edge the ant crosses:

Answer: 17. The ant crosses the long 15 edge, reaching it 15 × 3 over 8 = 5.625 units from the near end. For comparison, walking along an edge and then across a face costs 15 + √34 ≈ 20.83, and the straight line through the inside (which the ant can't use) is √259 ≈ 16.09.

The rule: for a box with dimensions ℓ ≥ w ≥ h, the shortest surface path between opposite corners is √ℓ² + (w + h)². Compare the 3D diagonal, which goes through the inside.

After folding, QM is the same length as QA. Call it y and look at triangle QDM.

Folding doesn't change lengths, so QM = QA = y, and QD = 12 − y. Triangle QDM has a right angle at D and DM = 6:

The crease is the perpendicular bisector of AM. In the same way it meets BC at P, 1.5 cm above B (solve 12² + t² = 6² + (12 − t)²). So the crease runs from (0, 7.5) to (12, 1.5):

Answer: AQ = 7.5 cm; the crease is 6√5 ≈ 13.42 cm. The key move, "a fold makes two lengths equal", solves almost every paper-folding problem.

The four corner triangles are identical right triangles with legs x and y. What are x + y and x² + y²?

Each side of the tilted square is the hypotenuse of a corner triangle, so x² + y² = 625. Each side of the big square is x + y = 31. Square the second equation:

Two numbers with sum 31 and product 168 solve t² − 31t + 168 = 0, so (t − 7)(t − 24) = 0.

Answer: 7 and 24. Check: 49 + 576 = 625. This tilted-square picture is also a proof of the theorem itself: the big square's area is the tilted square plus four triangles, (x + y)² = c² + 2xy, which simplifies to x² + y² = c². See the rearrangement proof.

Tangent segments from the same point to a circle are equal. At the right angle, the two radii and the legs make a square of side r.

The radii to the two legs are perpendicular to them, so at the right angle they form a square of side r. That leaves tangent lengths a − r and b − r from the other two vertices. From each of those vertices, the two tangent segments are equal, so the hypotenuse is made of those same two pieces:

For 20, 99, 101: r = 20 + 99 − 101 over 2 = 9. Check another way: area = 1 over 2 · 20 · 99 = 990, semi-perimeter = 110, and r = area ÷ semi-perimeter = 9.

Answer: r = 9. Bonus: for every Pythagorean triple the inradius is a whole number, because a + b and c are always both odd or both even.

By symmetry the center lies on the line through the middle of the square. Join the center to one of the corners the circle passes through.

The circle touches the top side at its midpoint, so the center is r below the top, which is 8 − r above the bottom. Joining the center to a bottom corner gives a right triangle with legs 4 and 8 − r and hypotenuse r:

Answer: r = 5. For a square of side s, the same steps give r = 5s over 8. The hidden triangle is a 3-4-5.

Put the corner at the origin. If the foot is at (p, 0) and the top at (0, q), what do you know about p² + q²? Where is the midpoint?

The ladder always has length 6, so p² + q² = 36. The midpoint is (p over 2, q over 2), and its distance from the corner is

Answer: a quarter circle of radius 3 m centered at the corner; the midpoint is always 3 m away, including when the foot is 2 m out. Check: then the top is √36 − 4 = 4√2 high, the midpoint is (1, 2√2), and √1 + 8 = 3.

A geometric view: the midpoint of the hypotenuse of a right triangle is the same distance from all three vertices. Other points on the ladder trace quarter-ellipses, an idea used in a drawing tool called the trammel of Archimedes. See the equation of a circle for why x² + y² = 9 is a circle.

First use the theorem to find how high each ladder reaches. Then use similar triangles: if the ladders reach heights A and B, the crossing height satisfies 1 over h = 1 over A + 1 over B.

Heights reached on the walls:

Let the crossing point be x from the left wall. Similar triangles along each ladder give h over A = x over 8 and h over B = 8 − x over 8. Adding them:

Answer: 30 over 7 ≈ 4.29 m, at x = 16 over 7 ≈ 2.29 m from the foot of the 17 m ladder. The alley's width cancels out of 1 over h = 1 over A + 1 over B.

The famous harder version runs the other way. Given the ladder lengths and the crossing height, find the width. That leads to a fourth-degree equation. For ladders of 20 and 30 crossing at a height of 8, the width is about 16.2, which in practice you find numerically.

α = 45°, so you need β + γ = 45°. Put the angles on a square grid with slopes 1 over 2 and 1 over 3, side by side at one point, and look for an isosceles right triangle.

The first line is the diagonal of a square, so α = 45°. For the others, a line with "rise 1, run 2" makes angle β with the horizontal, and "rise 1, run 3" makes angle γ. On a grid, draw OP to P(3, 1), above the horizontal, and OQ to Q(2, −1), below it. Then ∠POQ = β + γ.

Since OQ² + QP² = OP², the converse says the angle at Q is 90°. And OQ = QP, so triangle OQP is an isosceles right triangle, with base angles of 45°. So β + γ = 45° and α + β + γ = 90°.

Proved. As a numerical check, β ≈ 26.57° and γ ≈ 18.43°.

Draw lines through P parallel to the sides. Each distance from P to a corner is the hypotenuse of a right triangle whose legs are distances from P to two sides.

Let P be w from the left side, x from the right, y from the bottom and z from the top. Then

Both PA² + PC² and PB² + PD² equal w² + x² + y² + z². So for any point and any rectangle, PA² + PC² = PB² + PD², opposite corners paired:

Answer: PD = √31 ≈ 5.57. The name comes from the picture: the four right triangles look like the crosses on the Union Jack. One rectangle that works is about 8.08 by 8.66, but you never needed it.

Rotate the whole picture 60° about A so that B lands on C. Where does P go, and what triangle appears?

Rotate 60° about A, taking B to C. Point P goes to a new point P′ with AP′ = 3 and P′C = PB = 4. Because AP = AP′ and ∠PAP′ = 60°, triangle APP′ is equilateral, so PP′ = 3.

Now triangle PP′C has sides 3, 4 and 5, and 9 + 16 = 25, so the angle at P′ is 90°. So ∠AP′C = 60° + 90° = 150°.

To find AC, drop a perpendicular from C to the line AP′ extended past P′, with foot H. The angle at P′ outside the triangle is 30°, so triangle P′HC is a 30-60-90 triangle with hypotenuse 4: CH = 2 and P′H = 2√3. Then

Answer: s = √25 + 12√3 ≈ 6.77. A check with coordinates confirms that the point at distances 3, 4 and 5 lies inside this triangle. (With the law of cosines the last step is one line, but you don't need it.)

If the break is at height x, the part lying from the break to the ground has length 10 − x.

The standing part, the ground and the broken part form a right triangle with legs x and 3 and hypotenuse 10 − x:

Answer: 4.55 units (that is, 411 over 20). This problem, with these numbers, appears in the Nine Chapters on the Mathematical Art, a Chinese text compiled by about the 1st century CE, whose ninth chapter is devoted to right triangles. There the stalk is 1 zhang (10 chi) tall and the answer is given as 411 over 20 chi. Read more about the gougu theorem.

The obvious route (up 1, across the ceiling 30, down 11) is 42 ft. Try unfolding the room so the route crosses more faces: the ceiling, a side wall and the floor.

Unfold the room so that one end wall, the ceiling, a side wall, the floor and the other end wall lie flat in a chain, as in the net below. In this net the spider and the fly are 32 ft apart horizontally (1 + 30 + 1) and 24 ft apart vertically (6 + 12 + 6).

The straight line really does pass through those five faces in order. It crosses from the ceiling onto the side wall 7 ft along the room, and onto the floor at 23 ft. Other unfoldings are longer: the obvious route over the ceiling is 42 ft, and the route via the ceiling and one side wall is √37² + 17² = √1658 ≈ 40.72 ft.

Answer: 40 ft, two feet shorter than the obvious route, by crossing five of the six faces. This is the "spider and the fly", popularized by the English puzzle writer Henry Dudeney in the early 1900s. It is a 3-4-5 triangle in disguise: 32 : 24 : 40 = 4 : 3 : 5.