AMC 8, 2023, Problem Marathon

[Arisha Roy]

Always the smaller isosceles triangles are similar to the larger isosceles triangles.

The area of the gray part in the first triangle is \([A B C] \cdot\left(1-\left(\frac{11}{h}\right)^2\right)\).

Similarly, the area of the gray part in the second triangle is \([A B C] \cdot\left(\frac{h-5}{h}\right)^2\).

These areas are equal.

So, \(1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2\).

Simplifying it we get, \(10 h=146\)

So, \(h= 14.6\)

• This reply was modified 1 year, 5 months ago by Arisha Roy.

[Arisha Roy]

Problem 23

• This reply was modified 1 year, 5 months ago by Arisha Roy.

[Arisha Roy]

Solution:

There are 4 cases that the tiling will contain a large gray diamond in one of the smaller \(2 \times 2\) grids,

In each case, the number of ways to arrange the tiles other than the diamond = \(4^5\)

In all the 4 cases the number of ways will be = \(4 \cdot 4^5\)

The probability that the tiling will contain a large gray diamond in one of the smaller \(2 \times 2\) grids will be = \(\frac{4 \cdot 4^5}{4^9}=\frac{4^6}{4^9}=\frac{1}{4^3}=\frac{1}{64}\).

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