Olympiad Maths Prep

Track / Stage 8 / 7 of 180 #1707 of 2000

Problem 1707

IMO Shortlist mid-range; USAMO P2/P5
Geometry Difficulty 8.0 Prove it

It is given parallelogram ABCDABCD. On it's sides AB,BC,CD,DAAB, BC, CD, DA are chosen points E,F,G,HE, F, G, H such that area of EFGHEFGH is half of the area of ABCDABCD. Show that at least one of the quadrilaterals ABFHABFH and AEGDAEGD is parallelogram.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

1. Affine Transformation: We start by applying an affine transformation to map the parallelogram ABCDABCD to a unit square. This transformation preserves the ratio of areas and parallelism of lines. Let the vertices of the unit square be A(0,0)A(0,0), B(1,0)B(1,0), C(1,1)C(1,1), and D(0,1)D(0,1).

2. Define Points: Let the points E,F,G,HE, F, G, H be on the sides AB,BC,CD,DAAB, BC, CD, DA respectively. Define the coordinates of these points as follows:
- E=(a,0)E = (a, 0) where 0≤a≤10 \leq a \leq 1
- F=(1,b)F = (1, b) where 0≤b≤10 \leq b \leq 1
- G=(c,1)G = (c, 1) where 0≤c≤10 \leq c \leq 1
- H=(0,d)H = (0, d) where 0≤d≤10 \leq d \leq 1

3. Area Calculation: The area of quadrilateral EFGHEFGH can be calculated using the Shoelace formula:
Area=12∣a⋅b+1⋅c+c⋅d+0⋅0−(0⋅1+b⋅c+1⋅d+d⋅a)∣ \text{Area} = \frac{1}{2} \left| a \cdot b + 1 \cdot c + c \cdot d + 0 \cdot 0 - (0 \cdot 1 + b \cdot c + 1 \cdot d + d \cdot a) \right|
Simplifying, we get:
Area=12∣ab+c−bc−ad∣ \text{Area} = \frac{1}{2} \left| ab + c - bc - ad \right|

4. Given Condition: We are given that the area of EFGHEFGH is half the area of ABCDABCD. Since ABCDABCD is now a unit square, its area is 1. Therefore:
12∣ab+c−bc−ad∣=12 \frac{1}{2} \left| ab + c - bc - ad \right| = \frac{1}{2}
Simplifying, we get:
∣ab+c−bc−ad∣=1 \left| ab + c - bc - ad \right| = 1

5. Sum of Areas of Triangles: The area of the unit square is 1, and it is twice the sum of the areas of the four small right triangles near the square's angles. Therefore:
1=2(12ad+12(1−d)c+12(1−c)(1−b)+12b(1−a)) 1 = 2 \left( \frac{1}{2}ad + \frac{1}{2}(1-d)c + \frac{1}{2}(1-c)(1-b) + \frac{1}{2}b(1-a) \right)
Simplifying, we get:
1=ad+(1−d)c+(1−c)(1−b)+b(1−a) 1 = ad + (1-d)c + (1-c)(1-b) + b(1-a)

6. Simplify the Equation: Expanding and simplifying the equation:
1=ad+c−cd+1−c−b+bc+b−ab 1 = ad + c - cd + 1 - c - b + bc + b - ab
1=ad+1−cd−ab+bc 1 = ad + 1 - cd - ab + bc
1=ad+1−cd−ab+bc 1 = ad + 1 - cd - ab + bc
1=ad+1−cd−ab+bc 1 = ad + 1 - cd - ab + bc
1=ad+1−cd−ab+bc 1 = ad + 1 - cd - ab + bc
1=ad+1−cd−ab+bc 1 = ad + 1 - cd - ab + bc
1=ad+1−cd−ab+bc 1 = ad + 1 - cd - ab + bc
1=ad+1−cd−ab+bc 1 = ad + 1 - cd - ab + bc
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\[
1 = ad +

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.