The user-friendly version of this content is available here.

The following content is copyright (c) 2009-2013 by Goods of the Mind, LLC.

This essential trains for: Math Kangaroo 7-8, Math Kangaroo 9-10, AMC-10, AMC-12, AIME.

Often, contest problems are inspired from physics. Having some insight into the physics of the problem makes it easier to find a solution. Of course, the best would be to actually learn physics. The next best is to briefly go over the main ideas.

Problems inspired from Huygens' principle of wave propagation

Huygens stated a qualitative rule for finding the wavefront of a wave at a certain time if the wavefront at a previous time and the velocity of the wave are known.

A wavefront is the geometric locus of the points that are reached by a wave at a certain time.

Principle: *given the wavefront at time t, assume that each point that belongs to it becomes a source of the same type of oscillations as the main source. The wavefront at time t+Δt is the envelope (tangent) of the wavelets thus generated.*

This principle is used in contest problems that range from simple to extremely difficult:

Example 1:

Given a geometric figure, what is the set of points that can be reached by moving one unit away from the contour of the figure in any direction of the plane?

This is clearly an example of finding the wavefront after a fixed time if the existing wavefront is given. Simply find the envelope of the elementary wavelets:

Observe that the resulting contour is 'rounded' at the corners. One can easily imagine the resulting contour as being the same as the locus described by the center of the small circle as it revolves around the initial contour.

While revolving along a side of the polygon, the center describes a line segment. However, when it reaches a vertex, the circle pivots and the center describes a circular arc that connects smoothly with the segments on either side.

All in all, for a convex polygon it requires little experimentation to see that the new contour can also be thought of as a translation by one unit of each side of the polygon, parallel to itself plus the contour of a whole circle (as all the circular arcs involved add up to one whole rotation).

For example, if a polygon has a perimeter of length p, the length of the contour that encloses all the 1-unit translations of the points of the polygon is equal to:

Think: what happens if a similar question refers to concave polygons?

Example 2:

Another problem could be a variation of the problem posed by I.M.Gel'fand in his book "Trigonometry."

Gelfand's problem asks for the area of the set of points that a traveler can reach in a given time if he starts on a straight road and may decide at any moment to walk through the adjacent fields. The speed he can walk at on the road is different from the speed in the field and both speeds are constant.

Let us say the speed on the road is v1 = 5 mph and the one in the field is v2 = 2 mph. The traveler's decision to leave the road can happen at any time during the given time interval (let us say, 1 hr).

The figure shows the starting point as O and the road as a number line with origin at O. Obviously, the farthest points the traveler can reach in 1 hr on either side of O are 5 miles away: A and B.

At any arbitrary point P between O and the two ends of the segment AB, the set of points that can be reached in the adjacent fields is given by the shaded area:

Assume that, upon reaching the point P by walking on the road, the time remaining to walk until 1 hr has passed is t. Then, a tangent from B to the circle will form a right triangle with hypotenuse v1t and the leg corresponding to the radius of the circle v2t:

Since the point P has been chosen arbitrarily, we can easily find that the angle x does not depend on time:

This means that the tangent to any of these circles is a tangent to all such circles. We have thus found the envelope of the wavelets generated at each point on the wavefront AB.

Therefore, the set of points that can be reached from O consists of four congruent right triangular areas and two circular sectors:

Each triangle has area:

Example 3:

A variation of Gel'fand's problem has been used on AIME. The AIME problem states that a fire truck is placed at the origin of a rectangular system of coordinates. The two axes are roads and the rest of the plane is a field. The question is how large an area can this fire truck reach in 1 hour, given the different velocities for road and field.

Obviously, the shape of the area is going to depend on the ratio of the two speeds. Set up a theoretical framework and then try various values to see what the shape is:

Notice how in figure a the tangent lines intersect outside the circle forming a concave octagon, while in figure b the tangent lines intersect inside the circle and some circular arcs are also part of the final contour.

Problems inspired from Fermat's principle

Fermat stated that: *a light ray will travel between two points along the path that takes the shortest time.*

This helps understand the reflection and the refraction of light in qualitative terms. Also known as Snell's laws:

Reflection: *the angle of incidence equals the angle of reflection.*

In the figure, a particle starts at point A towards screen S. Assuming it will collide with the screen at point C, what will be its path after the collision? Its path forms an angle x with the perpendicular to S at C (also called 'normal' to S).

If the screen would be removed, then the particle would follow a linear path, eventually reaching point B. Fermat's principle states that, after the collision, the particle will eventually reach point B' which is symmetric to B with respect to the screen S, since the total length of its path from A to C to B' must not exceed the length of A--C--B.

This, of course, assumes that the collision is perfectly elastic and, therefore, the particle's speed (absolute value of velocity) does not change upon reflection.

This is not the case with refraction, where the speed changes at the boundary.

Example 1:

A ball bounces elastically at the boundaries of a rectangular box ABCD. If it starts at point C with a velocity that makes an angle of 60° with the wall CD, which wall will it collide against at the 12-th collision?

Since the elastic collision will always produce a path which is symmetric to the path in the absence of the collision, we can solve the problem graphically by 'eliminating the boundaries':

Now we can count the collisions and figure out which side they happen on.

Of course, the ratio width/length is important as well as the proper sequence in interpreting the grid lines (each vertical line can be either CD or BA, etc.).

Example 2:

Another variation on the same theme involves paths from one side of an orientable surface to the other side. The following problem has been a contest problem in Canada.

An ant is standing on the outside of a cylinder and wants to walk to a point inside it. What is the shortest path?

Since the cylinder can be flattened onto a plane without distortion, the shortest path must be a line segment.

In the figure, the outside surface is blue and the inside surface is yellow. After flipping the inside surface outward, the cylinder's lateral surface has been flattened into a rectangle.

The ant should cross the rim of the cylinder at point P.

The paths colored in green are obviously longer than the red one.