Intermediate geometry problems (ages 13 to 16)

Answer: C, 63

1. Put A = (0, 0), B = (12, 0), C = (12, 12), D = (0, 12). Then E = (12, 6) is the midpoint of BC and F = (3, 12) has DF = 3.
2. The square has area 12 × 12 = 144.
3. Corner triangle ABE: legs AB = 12, BE = 6, area ½ × 12 × 6 = 36.
4. Corner triangle ECF: legs EC = 6, CF = 12 − 3 = 9, area ½ × 6 × 9 = 27.
5. Corner triangle FDA: legs FD = 3, DA = 12, area ½ × 3 × 12 = 18.
6. Triangle AEF = 144 − 36 − 27 − 18 = 63 (C).

Why this works: A slanted triangle inside a rectangle is best found by subtraction: the pieces left over are right-angled and easy.

Where it leads: ‘Box and subtract’ works for any triangle with corners on a grid; it is the idea behind the shoelace formula.

Answer: B, 6

1. Hypotenuse: √(20² + 21²) = √(400 + 441) = √841 = 29.
2. Area: ½ × 20 × 21 = 210. Half-perimeter: s = (20 + 21 + 29)/2 = 35.
3. Joining the incentre to the vertices splits the triangle into three triangles of height r, so area = r × s.
4. r = 210 ÷ 35 = 6.
5. Check with the right-angle shortcut r = (a + b − c)/2 = (20 + 21 − 29)/2 = 6. ✓ Answer 6 (B).

Why this works: Splitting the triangle into three triangles from the incentre gives Area = r × s, where s is the half-perimeter. It works for every triangle.

Where it leads: For a right-angled triangle the inradius is also (a + b − c)/2: (20 + 21 − 29)/2 = 6.

Answer: 8

1. The perpendicular from the centre to a chord bisects it, making a right-angled triangle: (half-chord)² + d² = r².
2. Chord 6: half-chord 3, so d = √(25 − 9) = 4.
3. Chord 8: half-chord 4, so d = √(25 − 16) = 3.
4. Chords on opposite sides of the centre: 4 + 3 = 7 apart.
5. Chords on the same side: 4 − 3 = 1 apart.
6. Sum of the two possible distances: 7 + 1 = 8.

Why this works: The perpendicular from the centre bisects a chord, giving a right-angled triangle with the radius as hypotenuse. ‘Two possible answers’ is a cue to draw both configurations.

Where it leads: ‘Two possible configurations’ problems reward drawing both diagrams before calculating.

Answer: B, 8 + 8√2

1. Extend four sides of the octagon to form a square. The four cut-off corners are right-angled isosceles triangles, each with hypotenuse 2 (a side of the octagon).
2. A right-angled isosceles triangle with hypotenuse 2 has legs 2/√2 = √2.
3. The square’s side is √2 + 2 + √2 = 2 + 2√2, so its area is (2 + 2√2)² = 4 + 8√2 + 8 = 12 + 8√2.
4. The four corners have area 4 × ½ × √2 × √2 = 4.
5. Octagon: 12 + 8√2 − 4 = 8 + 8√2 (B), about 19.3.

Why this works: A regular octagon is a square with its corners snipped off. Building up to a simpler shape and subtracting is often easier than splitting into pieces.

Where it leads: A regular octagon of side s has area 2(1 + √2)s². Stop signs are regular octagons.

Answer: 8

1. Heron’s formula: half-perimeter s = (10 + 17 + 21)/2 = 24.
2. Area = √(s(s − a)(s − b)(s − c)) = √(24 × 14 × 7 × 3) = √7056 = 84.
3. For each side, altitude = 2 × area ÷ side, so the longest side has the shortest altitude.
4. Altitude to 21: 168 ÷ 21 = 8. (To 17: about 9.9; to 10: 16.8.)
5. Shortest altitude: 8.

Why this works: Every altitude times its base gives twice the same area, so altitudes are inversely proportional to the sides they meet.

Where it leads: Triangles with whole-number sides and area are called Heronian; 10, 17, 21 is made from two Pythagorean triangles (6-8-10 and 8-15-17) glued along the height 8.

Answer: D, 1/8

1. The water fills a cone at the bottom (the point), with the same shape as the whole cone.
2. Its height is half the full height, so every length (height and radius) is half: the scale factor is k = 1/2.
3. Volumes of similar solids scale by k³: (1/2)³ = 1/8.
4. Check with the formula: ⅓π(r/2)²(h/2) = (1/8) × ⅓πr²h. ✓
5. The water is 1/8 of the volume (D).

Why this works: For similar solids, lengths scale by k, areas by k² and volumes by k³. Half the height holds only an eighth of the volume.

Where it leads: Scaling lengths by k scales volumes by k³: so to half-fill a cone by volume you need about 79% of the height.

Answer: 16

1. Lattice points on the circle are integer pairs (x, y) with x² + y² = 65.
2. Squares up to 65: 0, 1, 4, 9, 16, 25, 36, 49, 64. Pairs adding to 65: 1 + 64 and 16 + 49 (65 − 0, 65 − 4, 65 − 9, 65 − 25, 65 − 36 are not squares).
3. So {|x|, |y|} = {1, 8} or {4, 7}.
4. Each pair gives 2 orders and 4 sign patterns: 8 points, e.g. (1, 8), (8, 1), (−1, 8), ….
5. Total: 8 + 8 = 16.

Why this works: Lattice points on x² + y² = n come from ways to write n as a sum of two squares; symmetry (swaps and signs) multiplies each one by up to 8.

Where it leads: 65 = 5 × 13 is a product of two primes of the form 4k + 1, which is why it has two different representations as a sum of two squares.

Answer: 84

1. Let the height onto the side of length 14 be h, splitting it into x and 14 − x.
2. Pythagoras twice: h² = 13² − x² = 15² − (14 − x)². Expanding: 169 − x² = 225 − 196 + 28x − x², so 28x = 140 and x = 5.
3. h = √(169 − 25) = 12.
4. Area = ½ × 14 × 12 = 84.

Why this works: Splitting a triangle by an altitude gives two right-angled triangles sharing the height, and equating the two expressions for h² finds where the foot lands.

Where it leads: Heron’s formula gives the same: s = 21, area = √(21 × 8 × 7 × 6) = 84. Triangles with whole-number sides and area are called Heronian.

Answer: B, 5

1. The points (0, 0), (8, 0), (0, 6) form a triangle with a right angle at the origin.
2. The angle in a semicircle is a right angle, so the hypotenuse from (8, 0) to (0, 6) is a diameter.
3. Its length is √(64 + 36) = 10, so the radius is 5 (B).

Why this works: Recognising the right angle uses the converse of ‘the angle in a semicircle is 90°’ and avoids solving equations for the centre.

Where it leads: For any triangle, the circumradius is R = abc/(4 × area). For a right-angled triangle that gives half the hypotenuse.

Answer: B, 4√6

1. The radii to the two touching points are both perpendicular to the tangent, so the centres and touching points form a right trapezium.
2. Draw a line from the smaller centre parallel to the tangent. It makes a right-angled triangle with hypotenuse 10 (between centres) and one side 5 − 3 = 2.
3. The other side equals the tangent length: √(10² − 2²) = √96 = 4√6 (B), about 9.8.

Why this works: Tangents meet radii at right angles, so sliding the tangent to a centre creates a right-angled triangle.

Where it leads: For the ‘crossing’ tangent (circles on opposite sides) use 5 + 3 instead: √(100 − 64) = 6. These lengths matter in belt-and-pulley design.

Answer: 72 cm²

1. The square’s diagonals pass through the centre, so each is a diameter: 12 cm.
2. The square can be split along a diagonal into two triangles, each with base 12 and height 6, or use area = d²/2.
3. Area = 144/2 = 72 cm².

Why this works: Area = d²/2 for any square (and half the product of the diagonals for any rhombus or kite) avoids finding the side.

Where it leads: The square fills 72/(36π) = 2/π ≈ 64% of the circle. A regular n-gon in the circle fills more and more as n grows.

Answer: B, √3

1. (√3/4)s² = 9√3 gives s² = 36, so s = 6.
2. The height is (√3/2) × 6 = 3√3.
3. In an equilateral triangle the incentre lies one third of the way up the height, so r = 3√3/3 = √3 (B).

Why this works: In an equilateral triangle the centres coincide, and the medians meet a third of the way up, which gives the inradius from the height.

Where it leads: For any triangle, area = r × s where s is the semi-perimeter: here 9√3 = r × 9. That formula is the quickest route.

Answer: C, 12π

1. Full circumference = 2π × 6 = 12π. The arc 4π is 1/3 of it.
2. So the sector is 1/3 of the circle.
3. Area = ⅓ × π × 6² = 12π (C).

Why this works: Arc length and sector area are the same fraction of the whole circle.

Where it leads: In general sector area = ½ × radius × arc length, exactly like a triangle with base the arc and height the radius.

Answer: B, (−3, 4)

1. Reflect in y = x: (3, 4) → (4, 3).
2. Rotate 90° anticlockwise: (x, y) → (−y, x), so (4, 3) → (−3, 4).
3. The final point is (−3, 4) (B).

Why this works: Writing each transformation as a rule on coordinates makes combining them mechanical.

Where it leads: The combination (x, y) → (−x, y) is itself a reflection, in the y-axis. A reflection followed by a rotation is always a reflection: this is how the symmetries of a shape form a group.

Answer: 65

1. 180(n − 2) = 1980, so n − 2 = 11 and n = 13.
2. Each corner joins by a diagonal to n − 3 = 10 others; 13 × 10 counts each diagonal twice.
3. Diagonals: 13 × 10 / 2 = 65.

Why this works: Two standard facts chained together: the angle sum identifies the polygon, the double-counting formula counts its diagonals.

Where it leads: Splitting an n-gon into triangles by diagonals from one corner gives n − 2 triangles, which is where 180(n − 2) comes from.

Answer: 3

1. The vector (3, 4) is perpendicular to the line 3x + 4y = 26.
2. Points (1 + 3t, 2 + 4t) lie on the perpendicular through (1, 2). On the line: 3 + 9t + 8 + 16t = 26, so 25t = 15 and t = 3/5.
3. The distance is t × |(3, 4)| = (3/5) × 5 = 3.

Why this works: The coefficients of x and y in ax + by = c give a perpendicular direction, so the foot of the perpendicular is one short calculation away.

Where it leads: This gives the formula |ax₀ + by₀ − c| / √(a² + b²), which extends to planes in 3D.

Answer: 10 cm

1. Cut the tin by a vertical plane through its axis: the cross-section is a rectangle 6 cm wide (the diameter) and 8 cm tall.
2. The longest segment inside is the rectangle’s diagonal.
3. √(6² + 8²) = 10 cm.

Why this works: Choosing the right cross-section turns a 3D question into Pythagoras in a rectangle.

Where it leads: For a cuboid a × b × c, the longest rod is √(a² + b² + c²): Pythagoras applied twice.

Answer: A, 2√13

1. Cosine rule: BC² = 8² + 6² − 2 × 8 × 6 × cos 60° = 64 + 36 − 48 = 52.
2. (Without the cosine rule: the foot of the perpendicular from C is 3 from A, the height is 3√3, and BC² = 5² + 27 = 52.)
3. BC = √52 = 2√13 (A), about 7.2.

Why this works: With two sides and the angle between them, the cosine rule gives the third side directly; the perpendicular method shows why.

Where it leads: With 60° the cosine rule becomes c² = a² − ab + b². Triangles with a 60° angle and whole-number sides (like 3, 7, 8) are the ‘Eisenstein triples’.

Answer: 22

1. Split along the diagonal from (0, 0) to (4, 6).
2. Triangle (0, 0), (5, 1), (4, 6): ½|5 × 6 − 1 × 4| = ½ × 26 = 13.
3. Triangle (0, 0), (4, 6), (−1, 3): ½|4 × 3 − 6 × (−1)| = ½ × 18 = 9.
4. Total area: 13 + 9 = 22.

Why this works: A triangle with a corner at the origin and others at (a, b), (c, d) has area ½|ad − bc|, so any polygon can be done as a sum of these.

Where it leads: Summing these pieces all the way round is the shoelace formula. It works for any polygon whose sides do not cross.

Answer: 28

1. The trapezium is symmetric (equal slanting sides). Perpendiculars from the short side cut off two right-angled triangles with base 3 and hypotenuse 5.
2. So the height is √(25 − 9) = 4.
3. Area = ½(10 + 4) × 4 = 28.

Why this works: Symmetry tells you how the extra 6 units of the long side are split, and Pythagoras gives the height.

Where it leads: If the two slanting sides are different, set the overhangs as x and 6 − x and use Pythagoras twice: the same method as for the 13-14-15 triangle.

Answer: 12

1. Triangles ABD and ADC share the height from A, so their areas are in the ratio BD : DC = 2 : 3. Area ABD = 2/5 × 60 = 24.
2. Triangles ABE and EBD share the height from B, and AE = ED, so they have equal area.
3. Area BED = 24 / 2 = 12.

Why this works: ‘Same height, so areas are in the ratio of the bases’ lets you pass areas along a chain of triangles without any lengths.

Where it leads: This area-ratio method proves Ceva’s theorem and finds where medians meet. It is also the idea behind barycentric coordinates.

Answer: B, π/6

1. If the cube has side 2r, the ball has radius r.
2. Ball: (4/3)πr³. Cube: 8r³.
3. Ratio: (4/3)π / 8 = π/6 (B), about 52%.

Why this works: Writing both volumes in terms of the same length makes the ratio independent of size.

Where it leads: In 2D the circle fills π/4 ≈ 79% of its square; in 3D only 52%; in 10 dimensions the ball fills about 0.25%. High-dimensional cubes are mostly corners.

Answer: 35°

1. Angle AOB = 70° and angle ACB both stand on the minor arc AB.
2. The angle at the centre is twice the angle at the circumference on the same arc.
3. So angle ACB = 70° / 2 = 35°, wherever C is on the major arc.

Why this works: The inscribed angle theorem means every point on the major arc sees the chord at the same angle.

Where it leads: A point D on the minor arc sees AB at 180° − 35° = 145°: opposite angles of a cyclic quadrilateral add to 180°.

Answer: C, 25

1. The condition |x| + |y| ≤ 3 describes a square standing on one corner, symmetric about both axes, so count row by row.
2. Row by row: y = 0 gives 7 points; y = ±1 gives 5 each; y = ±2 gives 3 each; y = ±3 gives 1 each.
3. Total: 7 + 2(5 + 3 + 1) = 7 + 18 = 25 (C).

Why this works: Counting along rows uses the shape of the diamond: each step away from the centre loses one point at each end.

Where it leads: For |x| + |y| ≤ n the count is n² + (n + 1)², the ‘centred square numbers’. In 3D, |x| + |y| + |z| ≤ n gives octahedral numbers.

Answer: C, 3 : 2

1. Same perimeter: hexagon side s, triangle side 2s (6s = 3 × 2s).
2. The hexagon splits into 6 small equilateral triangles of side s.
3. The big triangle of side 2s splits into 4 such small triangles.
4. Ratio 6 : 4 = 3 : 2 (C).

Why this works: Cutting both shapes into the same small triangles compares areas by counting, with no square roots needed.

Where it leads: For a fixed perimeter, regular polygons with more sides have more area; the circle beats them all (the isoperimetric inequality).

Answer: 12 cm

1. The radius to the point of contact meets the tangent at a right angle.
2. So the centre, the contact point and P form a right-angled triangle with hypotenuse 13 and one side 5.
3. Tangent length = √(169 − 25) = 12 cm.

Why this works: The right angle between radius and tangent turns tangent lengths into Pythagoras.

Where it leads: The square of the tangent length, 144 = 13² − 5², is the ‘power’ of P with respect to the circle; it also equals PA × PB for any line through P cutting the circle at A and B.

Answer: 7.5 cm

1. Points on the crease are equidistant from the two corners that meet, so the crease lies along the perpendicular bisector of the diagonal joining them. The diagonal has length √(64 + 36) = 10.
2. Put the rectangle with corners (0, 0) and (8, 6). The perpendicular bisector passes through the centre (4, 3) with gradient −8/6 = −4/3.
3. It meets the bottom edge y = 0 at x = 4 + 3 × 3/4 = 6.25 and the top edge y = 6 at x = 1.75.
4. Crease length = √((6.25 − 1.75)² + 6²) = √(20.25 + 36) = √56.25 = 7.5 cm.

Why this works: A fold maps one corner onto the other, so the crease is their perpendicular bisector: geometry of reflections, not trial and error.

Where it leads: For an a by b sheet (a ≥ b) the crease is b√(a² + b²)/a. Paper folding can even trisect angles, which ruler and compasses cannot.

Answer: B, 36√2

1. The base diagonal is 6√2, so each base corner is 3√2 from the centre.
2. A slanting edge (6) is the hypotenuse of a right-angled triangle with sides 3√2 and the height h: h² = 36 − 18 = 18, so h = 3√2.
3. Volume = ⅓ × base area × height = ⅓ × 36 × 3√2 = 36√2 (B), about 50.9.

Why this works: Finding a right-angled triangle inside the solid (height, half-diagonal, edge) is the standard way into 3D measurement.

Where it leads: Two of these pyramids glued base to base make a regular octahedron, with volume 72√2 for edge 6.