Tiêu đề 49 Rotation of Figures.pdf ✅

MSTC 49: Rotation of Figures This portion discusses some possible situations where rotation of figures is necessary to reduce the number of variables and steps needed to solve a geometry problem. The situations highlighted key points in some of the past board exams. 1. Point inside an equilateral with distances from vertices [FLOW OF SOLUTION] • Rotate a side about a corner until it coincides with another side. • Check for triangle PBP’. Due to rotations, it shows that PBP′ is equilateral. Therefore, PP ′ = b ∠BPP ′ = 60° • Consider triangle PCP′. Using cosine law, ∠CPP′ can be solved, ∠CPP ′ = θ θ = cos−1 b 2 + c 2 − a 2 2bc

cos θ = (b√2) 2 + c 2 − a 2 2(b√2)c • Consider triangle CPD. Let s be the side length of the square s 2 = b 2 + c 2 − 2bc cos(θ + 45°) The side length of the square is now solvable. 3. Point inside an isosceles right triangle with distances from 3 vertices • Rotate the figure around the vertex with the right angle. • Consider triangle PAP′. It shows that it is an isosceles triangle. Hence, PP ′ = b√2 ∠APP ′ = 45° • Consider triangle CPP′. Set ∠CPP ′ = θ
cos θ = c 2 + (b√2) 2 − a 2 2cb√2 • Consider triangle APC. Let s be the length of a leg in the triangle. s 2 = b 2 + c 2 − 2bc cos(θ + 45°) The leg of the triangle is now solvable.