Maths Olympiad Prep

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, 2013

Geometry Difficulty 2.8 Junior Find the answer Canada

In right-angled, isosceles triangle FGHFGH, FH=8FH=\sqrt{8}. Arc FHFH is part of the circumference of a circle with centre GG and radius GHGH, as shown.

The area of the shaded region is

Pick one

Solution

We use labels, mm and nn, in the fourth row of the grid, as shown.

[[IMAGE0]]

Then, 10,m,36,n10,m,36,n are four terms of an arithmetic sequence.

Since 10 and 36 are two terms apart in this sequence, and their difference is 3610=2636-10=26, the constant added to one term to obtain the next term in the fourth row is 262\frac{26}{2} or 13.

That is, m=10+13=23m=10+13=23, and n=36+13=49n=36+13=49.

(We confirm that the terms 10,23,36,4910,23,36,49 do form an arithmetic sequence.)

In the fourth column, 25 and nn (which equals 49) are two terms apart in this sequence, and their difference is 4925=2449-25=24. Thus, the constant added to one term to obtain the next term in the fourth column is 242\frac{24}{2} or 12.

That is, x=25+12=37x=25+12=37 (or x=4912=37x=49-12=37).
The completed grid is as shown.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.