Senior geometry problems (ages 16 to 18)

Answer: B, 5/4

1. Put B = (−6, 0), C = (6, 0) and A = (0, 8): AB = AC = √(36 + 64) = 10. Both centres lie on the axis x = 0.
2. Area = ½ × 12 × 8 = 48 and half-perimeter s = (10 + 10 + 12)/2 = 16, so the inradius r = 48/16 = 3: the incentre is (0, 3).
3. Circumradius R = abc/(4 × area) = (10 × 10 × 12)/192 = 25/4. The circumcentre is 25/4 below A: (0, 8 − 25/4) = (0, 7/4).
4. Distance: 3 − 7/4 = 5/4.
5. Check with Euler’s formula OI² = R(R − 2r) = (25/4)(1/4) = 25/16, so OI = 5/4. ✓ Answer 5/4 (B).

Why this works: Symmetry puts both centres on one line, so the distance is a subtraction. Euler’s formula OI² = R(R − 2r) gives an independent check.

Where it leads: Euler’s formula OI² = R(R − 2r) gives the same: (25/4)(25/4 − 6) = 25/16.

Answer: B, 3 − 2√2

1. Put the right angle at the origin with the legs along the axes (lengths 3 and 4).
2. Inradius of a right triangle: r = (3 + 4 − 5)/2 = 1, so the incircle has centre (1, 1) and radius 1.
3. A circle touching both axes has centre (t, t) and radius t.
4. It touches the incircle from outside, so the distance between centres is 1 + t: √2 (1 − t) = 1 + t.
5. Solve: t(1 + √2) = √2 − 1, so t = (√2 − 1)/(√2 + 1) = (√2 − 1)² = 3 − 2√2.
6. The radius is 3 − 2√2 ≈ 0.17 (B).

Why this works: Circles tangent to both arms of a right angle have centres on the bisector y = x, so tangency becomes one distance equation. Rationalising the denominator gives the neat form.

Where it leads: Chains of circles inscribed in a corner shrink geometrically: each is (3 − 2√2) times the last.

Answer: A, 32 − 8π

1. |x| + |y| ≤ 4 is a square turned on its corner, with vertices (±4, 0) and (0, ±4); its diagonals are 8, so its area is 8 × 8 ÷ 2 = 32.
2. x² + y² ≥ 8 is everything outside the circle of radius √8 = 2√2 about the origin.
3. The distance from the origin to the side x + y = 4 is 4/√2 = 2√2, exactly the radius: the circle touches each side and lies inside the square.
4. So the region is the square with the whole disc removed: area 32 − π × 8.
5. Answer: 32 − 8π ≈ 6.9 (A).

Why this works: Recognising |x| + |y| ≤ c as a rotated square and comparing the distance to its sides with the radius tells you whether the shapes overlap before any integration.

Where it leads: Area of the square (32) minus area of the inscribed circle (8π): the circle fits exactly inside the tilted square.

Answer: B, 18√2

1. The base is an equilateral triangle of side 6: area (√3/4) × 36 = 9√3.
2. The apex is directly above the centre of the base. The centre is 2/3 of the way along a median; the median has length 3√3, so the centre is 2√3 from each base vertex.
3. Height, by Pythagoras on an edge: h = √(6² − (2√3)²) = √(36 − 12) = √24 = 2√6.
4. Volume = ⅓ × base × height = ⅓ × 9√3 × 2√6 = 6√18 = 18√2.
5. Answer: 18√2 ≈ 25.5 (B).

Why this works: Volume = ⅓ × base × height for any pyramid; the height comes from Pythagoras once you know the apex is above the centroid of the base.

Where it leads: A regular tetrahedron of edge a has volume a³/(6√2); it fills exactly a third of the cube it can be inscribed in.

Answer: C, 24√2/7

1. The bisector splits the right angle at C into two 45° angles. Let CD = d.
2. Triangle ACD has sides 6 and d with the 45° angle between them: area ½ × 6 × d × sin 45°.
3. Triangle BCD: area ½ × 8 × d × sin 45°.
4. Together they make triangle ABC, area ½ × 6 × 8 = 24: ½ × 14 × d × (√2/2) = 24, i.e. (7√2/2) d = 24.
5. d = 48/(7√2) = 48√2/14 = 24√2/7.
6. CD = 24√2/7 ≈ 4.85 (C).

Why this works: Adding areas with the ½ab sin C formula is a quick route to any angle-bisector length: here d = 2ab cos(C/2)/(a + b).

Where it leads: In general the bisector of the right angle has length √2ab/(a + b): a harmonic-mean formula.

Answer: B, √74

1. The shortest route on the surface crosses two faces; unfolding those faces flat turns it into a straight line, the diagonal of a rectangle.
2. Unfolding pairs the edges in three ways: the rectangle is (a + b) by c for any split of the edges 3, 4, 5.
3. (3 + 4) by 5: √(49 + 25) = √74.
4. (3 + 5) by 4: √(64 + 16) = √80. (4 + 5) by 3: √(81 + 9) = √90.
5. The shortest is √74 ≈ 8.6. (√50, the space diagonal, would go through the inside of the box.)
6. Answer: √74 (B).

Why this works: Shortest paths on a surface become straight lines once the surface is flattened. Try every way of flattening and keep the best: here, add the two shortest edges.

Where it leads: Shortest paths on surfaces are straight lines in some unfolding; on a box, the best unfolding pairs the two smaller dimensions.

Answer: 31

1. Strictly inside means x²/16 + y²/9 < 1, so |x| ≤ 3 and |y| ≤ 2 (x = ±4 or y = ±3 give at least 1).
2. For each x, we need y² < 9(1 − x²/16).
3. x = 0: y² < 9, y = −2 to 2 (5). x = ±1: y² < 8.44 (5 each). x = ±2: y² < 6.75 (5 each).
4. x = ±3: y² < 9 × 7/16 ≈ 3.94, so y = −1, 0, 1 (3 each).
5. Total: 5 + 10 + 10 + 6 = 31.

Why this works: Counting lattice points column by column only needs the height of the curve at each integer x. Be careful at the edges: ‘strictly inside’ excludes (0, ±3) and (±4, 0).

Where it leads: Counting lattice points inside an ellipse approximates its area (12π ≈ 37.7); the error term is a famous research problem (Gauss circle problem).

Answer: 65/8

1. Area: split along the altitude to the side 14, which is 12 (5-12-13 and 9-12-15 triangles), so the area is ½ × 14 × 12 = 84.
2. R = abc / (4 × area) = (13 × 14 × 15) / 336 = 2730 / 336.
3. R = 65/8 = 8.125.

Why this works: R = abc/(4K) combines the sine rule (a = 2R sin A) with area = ½bc sin A.

Where it leads: The inradius is K/s = 84/21 = 4. Euler’s formula OI² = R(R − 2r) links the two centres: here OI² = 65/64.

Answer: 9

1. Corners: y = x meets y = 0 at (0, 0); y = 6 − x meets y = 0 at (6, 0); y = x meets y = 6 − x at (3, 3).
2. Base along y = 0 has length 6; height 3.
3. Area = ½ × 6 × 3 = 9.

Why this works: Find the vertices from pairs of lines, then use the side on an axis as the base.

Where it leads: The two slanted lines are perpendicular (gradients 1 and −1), so this is a right-angled isosceles triangle: area also = ½(3√2)².

Answer: 8

1. Angles CAB and CDB stand on the same arc CB, so they are equal; vertical angles at P are equal. So triangles APC and DPB are similar.
2. From the similarity, AP/DP = CP/BP, i.e. AP × PB = CP × PD.
3. 4 × 6 = 3 × PD, so PD = 8.

Why this works: Equal angles on the same arc give similar triangles, which turn into the product rule for crossing chords.

Where it leads: AP × PB is the same for every chord through P: it is the ‘power of the point’, equal to R² − OP².

Answer: B, 1/3

1. Take edge 2. Let M be the midpoint of edge AB. In faces ABC and ABD, MC and MD are perpendicular to AB, each of length √3.
2. The angle between the faces is angle CMD. CD = 2.
3. Cosine rule: 4 = 3 + 3 − 2 × 3 cosθ, so cosθ = 1/3 (B), θ ≈ 70.5°.

Why this works: The angle between two planes is measured with two lines, one in each plane, both perpendicular to their common edge.

Where it leads: Because 70.5° does not divide 360°, regular tetrahedra cannot fill space around an edge, a fact Aristotle got wrong.

Answer: B, √2

1. Put the corners at alternate corners of a cube: (1, 1, 1), (1, −1, −1), (−1, 1, −1), (−1, −1, 1). Edges have length 2√2.
2. Opposite edges are diagonals of opposite faces of the cube (e.g. on the faces x = 1 and x = −1), and they are perpendicular. The distance between them is the cube’s side, 2.
3. Scaling the edge from 2√2 down to 2 multiplies distances by 1/√2: the distance is √2 (B).

Why this works: Embedding the tetrahedron in a cube makes its symmetry visible and turns a skew-line distance into a cube side.

Where it leads: The common perpendicular joins the midpoints of the two edges. The three such segments of a tetrahedron meet at its centre, at right angles to each other.

Answer: B, √6

1. AB = (−1, 2, 0) and AC = (−1, 0, 2).
2. AB × AC = (2 × 2 − 0, 0 × (−1) − (−1) × 2, 0 − 2 × (−1)) = (4, 2, 2).
3. |(4, 2, 2)| = √24 = 2√6, so the area is √6 (B).

Why this works: The vector product’s length is the area of the parallelogram on two sides, so half of it is the triangle’s area.

Where it leads: ‘3D Pythagoras’: the square of this area equals the sum of the squares of its shadows on the three coordinate planes: 1 + 1 + 4 = 6. (de Gua’s theorem.)

Answer: 25/4

1. Put A = (0, 0), B = (10, 0), C = (10, 10), D = (0, 10). PA = PB means P = (5, y).
2. PA² = 25 + y²; distance to CD is 10 − y.
3. 25 + y² = (10 − y)² = 100 − 20y + y², so y = 15/4.
4. The distance is 10 − 15/4 = 25/4.

Why this works: Each ‘equal distance’ condition is a line or curve; coordinates turn their intersection into an equation.

Where it leads: P is the centre of the circle through A and B that touches CD. Points equidistant from a point and a line lie on a parabola: P is where that parabola meets the bisector.

Answer: 7

1. Let M be the midpoint of BC, so BM = MC = 4.
2. Apollonius’ theorem (cosine rule in triangles ABM and ACM, whose angles at M add to 180°): 49 + 81 = 2AM² + 2 × 16.
3. 2AM² = 98, AM = 7.

Why this works: The cosine rule in two triangles sharing the median, with supplementary angles at M, makes the cosines cancel.

Where it leads: In vectors: |b + c|² + |b − c|² = 2|b|² + 2|c|², the parallelogram law again.