I’ve been thoroughly enjoying Paul Zeitz’s book “The Art and Craft of Problem Solving“.

One of the early problems in the book is from 1994 Putnam math competition, but surprisingly easy once you see that it can be scaled into a much simpler question.

The problem :

**Find the positive value of ****m**** such that the area enclosed by the ellipse ****x^2/9 + y^2=1****, the x-axis, and the line ****y=2x/3**** is equal to the area in the first quadrant enclosed by the ellipse ****x^2/9 + y^2=1****, the y-axis, and the line ****y=mx****.**

Heres a drawing of the areas mentioned using GeoGebra, with each area in separate quadrants for comparison, with an approximate area. The Qn asks what is the slope of the line through point F [or -ve slope of line through Q here].

Problem as stated, with areas colored

Well its a bit hard to pick the point Q [ the slope -m ] so that these colored regions have the same area…

… but then you realize that the whole problem becomes easy if you simply scale the ellipse back to the unit circle.

To do this, x is scaled back by 3x, so areas become 1/3 of what they were [y remains the same]. So the problem now looks like this :

“]

scaled back to the unit circle : areas are x 1/3, slopes x 3

So the line that gives the same area is y=x/2. When this is scaled back to the original ellipse, the slope gets divided by 3; so the line we want is y=x/6.

Quite surprised to see this in a Putnam, but it does show a really common motif in math, namely :

**Transform to a simpler domain, solve it there, then transform the solution back**.

The astute reader will have noticed that I cheated : the areas of the colored regions in the top picture are of course ~1.65 or 3 x area of regions in the second diagram.

The areas marked were calculated by rough approximation with a polygon and the GeoGebra area function. [ The ellipse itself was drawn to fit using foci - I wasn't sure how to edit the formula in GeoGebra to make it exact. ]

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