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Geometry Difficulty 6.6 National olympiad Find the answer

A point (x,y)(x,y) in the first quadrant lies on a line with intercepts (a,0)(a,0) and (0,b)(0,b), with a,b>0a,b > 0. Rectangle MM has vertices (0,0)(0,0), (x,0)(x,0), (x,y)(x,y), and (0,y)(0,y), while rectangle NN has vertices (x,y)(x,y), (x,b)(x,b), (a,b)(a,b), and (a,y)(a,y). What is the ratio of the area of MM to that of NN?

[i]Proposed by Eugene Chen[/i]

A number or a short expression. Spacing and $ signs are ignored.

Solution

1. Determine the equation of the line:
The line passes through the intercepts (a,0)(a,0) and (0,b)(0,b). The equation of a line in standard form given intercepts aa and bb is:
xa+yb=1 \frac{x}{a} + \frac{y}{b} = 1
Multiplying through by abab to clear the denominators, we get:
bx+ay=ab bx + ay = ab

2. **Calculate the area of rectangle MM:**
Rectangle MM has vertices (0,0)(0,0), (x,0)(x,0), (x,y)(x,y), and (0,y)(0,y). The area of rectangle MM is:
Area of M=xy=xy \text{Area of } M = x \cdot y = xy

3. **Calculate the area of rectangle NN:**
Rectangle NN has vertices (x,y)(x,y), (x,b)(x,b), (a,b)(a,b), and (a,y)(a,y). The area of rectangle NN is:
Area of N=(ax)(by) \text{Area of } N = (a - x)(b - y)
Expanding this expression, we get:
Area of N=abaybx+xy \text{Area of } N = ab - ay - bx + xy

4. **Simplify the area of rectangle NN:**
From the equation of the line bx+ay=abbx + ay = ab, we can substitute abab with bx+aybx + ay:
Area of N=abaybx+xy=(bx+ay)aybx+xy \text{Area of } N = ab - ay - bx + xy = (bx + ay) - ay - bx + xy
Simplifying further, we see that the terms bxbx and ayay cancel out:
Area of N=xy \text{Area of } N = xy

5. Determine the ratio of the areas:
The ratio of the area of MM to the area of NN is:
Area of MArea of N=xyxy=1 \frac{\text{Area of } M}{\text{Area of } N} = \frac{xy}{xy} = 1

Conclusion:
1 \boxed{1}

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