Tiling (Proof Without Words II) of the Pythagorean Theorem — explained for a 16-year-old (Roger Nelson, medieval links, recreate in Desmos)

Goal: See why for a right triangle with legs a and b and hypotenuse c we have c^2 = a^2 + b^2 by a tiling (rearrangement) argument — the kind Roger Nelson presented as a "Proof Without Words II". We will (1) build the picture, (2) read off the areas, (3) note the medieval connections, and (4) show how to set it up in Desmos.

1. The picture (big idea).

   Start with a right triangle whose legs lie on the axes: vertices at (0,0), (a,0) and (0,b). Its hypotenuse length is c = sqrt(a^2 + b^2).

   Make four congruent copies of that triangle and place one in each corner of a square of side length (a + b) so that each triangle's legs run along the edges of the big square. When placed correctly the four triangles leave a central uncovered region — that region turns out to be a square whose side length equals c.

   So the big square is tiled by: four right triangles + the central c-by-c square.

2. Read off areas (the simple algebra).

   Area of the big square = (a + b)^2.

   Area of each right triangle = (1/2)ab, so area of four triangles = 4 * (1/2)ab = 2ab.

   Area of the central square = c^2.

   Because the big square is exactly the four triangles plus the central square,

   (a + b)^2 = 2ab + c^2.

   Expand the left side: a^2 + 2ab + b^2 = 2ab + c^2. Cancel 2ab from both sides to get

   c^2 = a^2 + b^2.

   That is the Pythagorean theorem. The argument is visual: the rearrangement (tiling) preserves area and forces the algebraic identity.

3. Why this is a "proof without words" and why tiling works.

   You don't need algebra to believe it once you see the picture: the big square's area is visibly the sum of the areas of the four identical triangles plus the central square. Area is preserved under the rearrangement, so the relationship must hold for all a and b. This geometry-to-algebra link is the power of tiling proofs.

4. Medieval context and Roger Nelson.

   The visual arrangement above is closely related to proofs given by medieval mathematicians — for example Bhaskara II (12th century, India) presented a very similar four-triangle square proof. Islamic geometric art and tiling traditions also explored how shapes fit together and often used area reasoning. Roger Nelson and other modern authors collected and published elegant visual proofs (often called "Proofs Without Words") — Nelson's "Tiling: Proof Without Words II" is a modern presentation of this same tiling philosophy.

5. How to build the figure in Desmos (step-by-step).

   Follow these steps to make an interactive diagram you can slide and explore.

   1. Make two sliders: a and b. (Start values: a = 3, b = 2.)
   2. Define the triangle’s vertices: A = (0,0), B = (a,0), C = (0,b). The triangle is polygon(A,B,C).
   3. Compute c as c = sqrt(a^2 + b^2) so you can display it and compare with the central square size.
   4. Draw the big square with corners at (0,0), (a + b,0), (a + b,a + b), (0,a + b).
   5. Place the four triangles in the four corners:
      • Lower-left: polygon((0,0),(a,0),(0,b)).
      • Lower-right: translate/rotate the triangle so its legs run along the bottom and right edges. In Desmos you can create it as polygon((a+b,0),(b,0),(a+b,-a)) if you prefer coordinate shifts — but the easiest is to reflect/rotate the original triangle about the appropriate edges using Desmos' point expressions or by constructing the three corner vertices explicitly.
      • Upper-right and upper-left: similarly place the other two triangles so each corner of the big square hosts one triangle, with legs aligned with the square sides.
   6. Shade the four triangles and the central region in different colors. The central region will appear as a square. Put a label on its side length: c = sqrt(a^2 + b^2).
   7. Turn on numerical labels or a text box showing areas: Big square area = (a + b)^2, Triangles total = 2ab, Central area = c^2. Slide a and b and watch the equality hold visually and numerically.

   Note: Desmos does not have an explicit polygon-rotation tool, but you can get the four triangle polygons by writing the corner coordinates directly in terms of a and b (common coordinates are simpler if you place the big square with lower-left at (0,0)). Many Desmos users also copy the triangle and apply simple translations/rotations with matrix formulas if they want an algebraic expression for each vertex.

6. Extra explorations (ideas):
   • Make a and b sliders allowing non-integer values (the tiles still fit and the central square side c updates correctly).
   • Compare this tiling to other Pythagorean proofs (e.g., Euclid's proof using similar triangles, or Bhaskara's moving-triangle proof).
   • Investigate tilings and patterns in medieval Islamic art: how repeated motifs encode geometric relationships.

Quick summary: The tiling (Proof Without Words II) places four congruent right triangles inside a square of side (a + b) leaving a central square of side c. Equating areas gives (a + b)^2 = 4*(1/2 ab) + c^2, which simplifies to c^2 = a^2 + b^2. This visual rearrangement is the heart of many medieval and modern geometric proofs of the Pythagorean theorem, and you can recreate and explore it interactively in Desmos by making sliders for a and b and drawing the polygons described above.