# On the Difference of Areas

The problem below that was originally proposed by Joseph Kennedy in *School Science and Mathematics* (**52**, 162, February 1952) appeared as a *quickie* in the *Mathematics Magazine* (Vol. 26, No. 5 (May - Jun., 1953), p. 287). It was also included in an eclectic collecton C. W. Trigg.

A circle of radius $15$ intersects another circle, radius $20,$ at right angles. What is the difference of areas of the non-overlapping portions?

### Reference

- C. W. Trigg,
*Mathematical Quickies*, Dover, 1985, #12

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A circle of radius $15$ intersects another circle, radius $20,$ at right angles. What is the difference of areas of the non-overlapping portions?

### Solution

The fact that the circles meet at right angles is a *red herring*. The solution is trivial and does not take that fact into account:

Let $X$ be the area of the intersection. Then the remaining portions of the two circles have the areas $\pi 20^{2} - X$ and $\pi 15^{2} - X$ with the difference

$(\pi 20^{2} - X) - (\pi 15^{2} - X) = \pi 20^{2} - \pi 15^{2},$

independent of $X.$

Indeed, it is not important that the two shapes be circles. The numerical answer will be the same for any two blobs of areas $\pi 20^2$ and $\pi 15^2.$

Let $A$ and $B$ be the areas of the two non-overlapping pieces such that the area of one shape is $A+X$ and the other $B+X.$ Then the difference in the areas of the shapes, $(A+X)-(B+X),$ is exactly the difference in the area of their non-overlapping parts:

$(A+X) - (B+X) = A-B.$

### What Is Red Herring

- On the Difference of Areas
- Area of the Union of Two Squares
- Circle through the Incenter
- Circle through the Incenter And Antiparallels
- Circle through the Circumcenter
- Inequality with Logarithms
- Breaking Chocolate Bars
- Circles through the Orthocenter
- 100 Grasshoppers on a Triangular Board
- Simultaneous Diameters in Concurrent Circles
- An Inequality from the 2015 Romanian TST
- Schur's Inequality
- Further Properties of Peculiar Circles
- Inequality with Csc And Sin
- Area Inequality in Trapezoid
- Triangles on HO
- From Angle Bisector to 120 degrees Angle
- A Case of Divergence
- An Inequality for the Cevians through Spieker Point via Brocard Angle
- An Inequality In Triangle and Without
- Problem 3 from the EGMO2017
- Mickey Might Be a Red Herring in the Mickey Mouse Theorem
- A Cyclic Inequality from the 6th IMO, 1964
- Three Complex Numbers Satisfy Fermat's Identity For Prime Powers
- Probability of Random Lines Crossing
- Planting Trees in a Row
- Two Colors - Three Points

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