-256x256.png)
These problems use the Pythagoras theorem and its consequences (perpendicularity criteria, projections, and equal‑tangent loci) to prove algebraic relations and describe geometric sets.
Before diving into geometric problems, it is useful to recall how the Pythagoras theorem and related algebraic identities appear from simple geometric dissections. Many classical proofs of Pythagoras are based on rearranging squares and triangles, or on counting areas. The following problems illustrate this connection.
1. Prove the following equalities:
a) (a + b)² = a² + 2ab + b²
b) (a − b)² = a² − 2ab + b²
c) (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc
d) a² − b² = (a + b)(a − b)
2. Given 4 right triangles with legs a, b and hypotenuse c, show that one can form a square of side a + b by adding:
a) one square of side c;
b) two squares of sides a and b.
3. Suppose a² + b² = c². Show how to dissect a square of side c into two squares of sides a and b (the number of parts should not depend on a and b).
4. In a rectangle drawn on squared paper, a diagonal is drawn. How many unit squares does it cross if the rectangle has size:
a) 7 × 13
b) 100 × 170
c) n × m
5. Suppose each matchstick has length 1 inch. Using 12 such matchsticks, form a figure with area 4 square inches.
6a. In triangle ABC, draw through A a line perpendicular to BC. Choose any point M on this line. Prove:
MB² − MC² = AB² − AC².
6b. In triangle ABC, suppose a point M in the plane satisfies
MB² − MC² = AB² − AC².
Prove that AM ⟂ BC.
7. From a point M inside triangle ABC, drop perpendiculars MP, MK, ME to sides AB, BC, CA respectively. Prove:
AP² + BK² + CE² = PB² + CK² + AE².
8. In an isosceles right triangle ABC, pick points M and K on hypotenuse AB (with K between M and B) so that ∠MCK = 45°. Prove that:
MK² = AM² + KB².
9. Find the locus of points from which the tangent lengths to two given circles are equal.
