# Practice Paper 1 Question 14

You must slice a square of side length \(b\) into 6 pieces with equal areas, using 3 lines that intersect each other inside the square, one of which is a diagonal. Where on the sides of the square, with respect to the nearest corner, should the other two lines cut? Give your answer in terms of \(b\).

## Related topics

## Warm-up Questions

- An isosceles triangle has base \(6\) and sides \(8\). What is its area?
- A square of side \(x\) sits on the base of an isosceles triangle with base \(x\) and area equal to the square's. How tall is the triangle?

## Hints

- Hint 1Consider why all three lines must intersect at the same point inside the square.
- Hint 2Consider why this point must be the centre of the square.
- Hint 3Why must the two non-diagonal lines be symmetric wrt the diagonal?
- Hint 4The configuration of the lines looks like this:
- Hint 5What are the areas of the different triangles in terms of \(b\) and \(x\)?

## Solution

The only way for the two free lines to yield 6 equal areas is if: all three lines must intersect at the same point, the intersection point must be the centre of the square, and the two free lines must be symmetric with respect to the given diagonal (*).

The four triangles sharing an edge with a diagonal will always have equal area, as do the other two, so we'll only concentrate on one of each. The first one has area \(\frac{b(b-x)}{4}\) (its height is \(b/2\)) and the second is found by subtracting from the half square the two bordering triangles, i.e. \(\frac{b^2}{2}-2\frac{b(b-x)b}{4}\). Equating we find \(x=\frac{b}{3}\) and hence \(b-x=\frac{2b}{3}\).

(*) Justification:

- If the lines do not all intersect at the same point inside the square then the only possibility is that the other two lines pass through some corner, otherwise there would be more than 6 resulting pieces. But in this case, the resulting areas can never be all equal. The intersection point must thus be inside the square.
- It must also be its centre. Any other point would give at least an area smaller than another (in particular, the triangle delimited by the smaller diagonal segment and the smaller line segment).
- The two lines must also intersect different square sides or else, looking at the triangles at the right/left of the diagonal, one of them will be larger than the other two.
- The two lines must be symmetric wrt both diagonals, otherwise the 1st and 3rd triangle from those on the right/left of the diagonal would have different areas.

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