8 original problems at the level of the first olympiad stage: triangles and circles: inradius, circumradius, chords, bisectors, lattice points. Each answer is a whole number. Try each one before opening the hints; the full solution explains why the method works and where the idea leads.
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A right-angled triangle has legs of length 20 and 21. Its inscribed circle has centre I. Find the square of the distance from I to the vertex at the right angle.
Hint
Find the hypotenuse, then the inradius of a right-angled triangle.
Second hint
For a right angle, r = (leg + leg − hypotenuse)/2, and I is r units from each leg.
Full worked solution
Answer: 72
Hypotenuse: √(202 + 212) = √841 = 29.
In a right-angled triangle the inradius is r = (20 + 21 − 29)/2 = 6 (the tangent lengths from the right-angle vertex are both r).
Put the right angle at the origin with the legs on the axes: I = (6, 6).
Distance squared: 62 + 62 = 72.
Why this works: Equal tangent lengths make the right-angle corner of the incircle a square of side r, so the inradius comes straight from the sides.
Where it leads: Equal tangents give the general inradius formula r = Area / s and lead on to the excircles.
Why this works: Area = abc/(4R) comes from Area = ½ab sin C together with the sine rule c = 2R sin C, so one area calculation gives the circumradius.
Where it leads: Triangles with integer sides and integer area (Heronian triangles) can be split into two right-angled triangles with rational sides, as this one splits into 6-8-10 and 15-8-17 along its altitude.
A rhombus has diagonals of length 24 and 10. A circle touches all four sides. Find 13 times its radius.
Hint
Find the side length using the half-diagonals.
Second hint
Area = (product of diagonals)/2, and also area = perimeter × r / 2.
Full worked solution
Answer: 60
The diagonals bisect each other at right angles, so a side is √(122 + 52) = 13.
Area = 24 × 10 / 2 = 120.
The rhombus splits into four triangles from the centre, each with height r: area = ½ × 52 × r = 26r.
26r = 120 gives r = 60/13, so 13r = 60.
Why this works: Any polygon with an incircle has area = semi-perimeter × inradius: cut it into triangles from the centre.
Where it leads: Quadrilaterals with an incircle (tangential quadrilaterals) are exactly those with AB + CD = BC + DA (Pitot’s theorem and its converse).
CBSE Math Revision