STEP Exam Practice Questions

STEP Practice Problems

These are authentic STEP exam questions that bring together coordinate geometry concepts in challenging, real-world scenarios. Work through them actively—attempt each part before consulting the solutions.

How to Use These Problems

• Read carefully: Each problem has multiple parts building in complexity
• Try first: Attempt each part before looking at the solution
• Check your reasoning: Compare your approach, not just your final answer
• Learn from mistakes: Understand why your approach did or didn’t work
• Return later: Revisit these problems after you’ve studied more material

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Question 1: Three Vertical Poles (STEP 1, 1993, Question 8)

Topic: 3D coordinate geometry, distance formula, angles between vectors

Problem

A sculpture consists of three vertical poles , , and with heights 2 metres, 3 metres, and 4 metres respectively.

(i) Prove that the intersection of the surface of a sphere with a plane is always a circle, a point, or the empty set. Prove that the intersection of the surfaces of two spheres with distinct centres is always a circle, a point, or the empty set.

[If you use coordinate geometry, a careful choice of origin and axes may help.]

(ii) Show that .

Vandals now shift the pole so that is unchanged and the pole is still straight but is vertically above with (in radians). Find the new angle in radians correct to four figures.

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Solution

Part (i): Geometric intersections

Sphere ∩ Plane:

Consider a sphere with centre and radius :

and a plane with unit normal :

Let be the perpendicular distance from the centre to the plane.

Any point in the intersection satisfies both equations. Writing , where is the foot of the perpendicular from to the plane and lies in the plane:

Therefore .

• If : , giving a circle of this radius.
• If : , giving a single point.
• If : no real exists, giving the empty set.

Intersection of Two Spheres:

For spheres with centres and radii :

Subtracting these equations gives:

This is the equation of a plane. The intersection of the two spherical surfaces is exactly the intersection of one sphere with a plane, which we’ve already shown is a circle, point, or empty set. ✓

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Part (ii): Finding angle

Initial configuration:

Place the ground points as an equilateral triangle:

The tops of the vertical poles are:

Computing the angle:

The vectors from to and from to are:

The dot product:

The magnitudes:

Therefore:

So ✓

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After pole is tilted:

The pole now has length 2, makes angle with , and its top is vertically above the line .

Horizontal displacement from :

Vertical rise:

New position:

The other poles remain: and

New vectors:

Dot product:

Magnitudes:

New angle:

Tip

This problem demonstrates how coordinate geometry, combined with vector calculations, lets you solve complex 3D geometry problems elegantly. The key steps: choose your coordinates carefully, compute vectors, use the dot product formula for angles.

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Question 2: Common Tangents to Circles (STEP 2, 2015, Question 8)

Topic: Circle geometry, homothety, collinearity, vector relationships

Problem

Two non-overlapping circles and have a common external tangent. The lines and are the two common external tangents, intersecting at point called the focus of and .

(i) Let and be the position vectors of the centres of and respectively, with radii and . Show that the position vector of is:

(ii) A third circle (with radius , where ) doesn’t overlap or . Let be the focus of and , and be the focus of and . Show that , , and are collinear.

(iii) Find a condition on , , and for to lie exactly halfway between and .

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Solution

Part (i): Position of focus P

A homothety (scaling transformation) with centre and ratio sends circle to circle . For this to map the circles:

The homothety formula gives:

Rearranging:

✓

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Part (ii): Collinearity of P, Q, R

By the same argument, the foci are:

Computing vectors from P:

where

Since both and are scalar multiples of the same vector , they are parallel. Therefore, , , and are collinear. ✓

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Part (iii): Q as midpoint of PR

For to be the midpoint of , we need:

From the scalar factors above:

This is the condition for to lie exactly halfway between and .

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Key Takeaways

• Homothety provides an elegant framework for circle problems
• Vector methods simplify 3D geometry and relationships
• Collinearity emerges naturally from parallel vectors
• Special conditions (like midpoint) constrain the geometry in specific ways

Note

These STEP problems require integrating multiple concepts: 3D coordinates, vectors, angle calculations, circle properties, and geometric relationships. Working through them strengthens your ability to see connections between different mathematical areas.