Japan Junior Mathematical Olympiad

Question

Consider a square consisting of 9 little squares of side length

1

arranged in

3

rows and

3

columns as in the figure below. Let us number the little squares as indicated in the figure.

①	②	③
⑧		④
⑦	⑥	⑤

Let

A

,

B

,

C

,

D

be points chosen from the interior (not on the boundary) of the squares

1

,

3

,

5

,

7

, respectively. Denote by

X

the sum of the areas of the intersections of the quadrilateral

A B C D

with the squares

1

,

3

,

5

and

7

. Denote also by

Y

the sum of the areas of the intersections of the quadrilateral

A B C D

with the squares

2

,

4

,

6

and

8

. Prove that the inequality

X < Y

holds.

Accepted Answer

We partition the intersection of the quadrilateral

A B C D

and the union of squares

1

,

3

,

5

,

7

into

8

triangles

T_{1}, T_{2}, \dots, T_{8}

as indicated in the figure below. Then,

X

equals the sum of the areas of the triangles

T_{1}, T_{2}, \dots, T_{8}

. We also partition the intersection of

A B C D

and the union of the squares

2

,

4

,

6

,

8

into

8

triangles

S_{1}, S_{2}, \dots, S_{8}

as indicated in the figure below. Then,

Y

equals the sum of the areas of the triangles

S_{1}, S_{2}, \dots, S_{8}

.

As can be seen from the figure, for each

i

(

1 \leq i \leq 8

), the triangles

T_{i}

and

S_{i}

have a side in common. If we consider this common side as the base for the triangles

T_{i}

and

S_{i}

, then the altitude of the triangle

T_{i}

is

< 1

, since by assumption

A

,

B

,

C

,

D

are points in the interior of the squares, while the altitude of

S_{i}

equals

1

. Therefore, for each

i

the area of

S_{i}

is greater than the area of

T_{i}

. Consequently,

Y

, the sum of the areas of

S_{1}, S_{2}, \dots, S_{8}

is greater than

X

, the sum of the areas of

T_{1}, T_{2}, \dots, T_{8}

.