Interactive geometry lesson
Where should a shortest reflected path touch?
A visual transformation removes the guesswork from a classic geometry problem. The board shows why the construction minimizes distance and why the incoming and outgoing angles match.
The short answer
Reflect the destination across the mirror, draw a straight line from the source to that reflected point, and use its intersection with the mirror. Reflection turns the broken path into a straight path, which is shortest.
What happens in the lesson
- 01
Make the family of possible paths explicit
The source and target sit on the same side of a mirror line. A path must touch the mirror somewhere, but many contact points are possible. Guessing several routes suggests a minimum exists without identifying or proving it.
- 02
Reflect the target across the boundary
The target is copied to the opposite side at the same perpendicular distance from the mirror. For every contact point on the mirror, the distance from that point to the target equals the distance to its reflected image.
- 03
Unfold the broken path
Replacing the second segment with its reflected counterpart converts any source–mirror–target route into a source–mirror–reflected-target route of equal total length. The two-segment optimization problem is now an ordinary point-to-point distance problem.
- 04
Use the straight-line minimum
The direct segment from the source to the reflected target is shorter than every bent alternative. Where it crosses the mirror is therefore the required contact point. Reflecting the final segment back gives the shortest physical route.
- 05
Recover the law of reflection
The unfolded route is straight, and reflection preserves angle. Folding it back makes the angle of incidence equal the angle of reflection. The construction proves both the minimum-distance claim and the equal-angle condition.
What this example is designed to teach
- Reflection preserves the distance from any point on the mirror to the target.
- Unfolding converts a broken route into an equal-length route to the reflected point.
- A straight segment supplies a global minimum, not merely a convincing-looking guess.
- Folding the construction back proves the equal-angle law of reflection.
Direct answers
Frequently asked questions
Why does reflecting the target reveal the shortest path?+
Reflection preserves distance to every point on the mirror. A route from the source to the mirror and then to the target therefore has the same length as a route from the source through the mirror to the reflected target. The shortest such route is straight.
How does the construction prove equal angles?+
The straight line to the reflected target crosses the mirror at the optimal point. Reflecting the second segment back preserves its angle with the mirror normal, so the incoming and outgoing angles are equal.
Is this only an optics problem?+
No. The same transformation solves many shortest broken-path problems involving a boundary, including routing, billiards, and contest geometry.