### Archimedes Triangle and Squaring of Parabola

Triangles formed by two tangents to a parabola and the chord connecting the points of tangency have been used by Archimedes in his study of the area of parabolic segments and bear his name. The chord is usually considered the base of the triangle.

### Archimedes' Lemma

Let, in an Archimedes triangle ABS, M be the midpoint of the base. Assume that SM crosses the parabola in O, and let the tangent to the parabola at O cross the sides of ABS in A_{1} and B_{1}, as shown. Then

The median MS is parallel to the axis of the parabola,

A

_{1}and B_{1}are the midpoints of the sides AS and BS of ΔABS,O is the midpoint of MS.

### Proof

In ΔA'B'F, lines AS and BS serve as perpendicular bisectors of sides A'F and B'F. They intersect at S, so that the perpendicular from S to the third side A'B' is bound to bisect the latter. In trapezoid ABB'A', this line is parallel to the bases AA' and BB' and passes through the midpoint of one side A'B'. It therefore passes through the midpoint M of other side as well.

Triangles AA

_{1}O and BB_{1}O are also Archemedean. So the first part applies: the medians from A_{1}and B_{1}are parallel to the axes of the parabola. But those medians serve as midlines in triangles AOS and BOS.As we just showed A

_{1}B_{1}is a midline in ΔABS. It therefore cuts in half any cevian from S, MS in particular.

### Theorem 2 (Squaring of parabola)

(See [Dörrie, pp. 239-242, Stein, pp.51-62].)

Parabola divides the area of an Archimedes triangle in ratio 2:1. In other words, the area of the parabolic segment AB equals 2/3 of the area of the Archimedes triangle ABS.

### Proof

Let area( ΔABS) = 1. Two thousand years before the invention of Calculus, Archimedes filled the parabolic segment with triangles, whose areas are easily arranged into a geometric series whose sum he already knew.

The first triangle in the series is the "inner triangle" ABO. From Lemma, _{1}B_{1}S) = 1/4. Therefore, _{1}O) + area( ΔBB_{1}O) = 1/4.

AA_{1}O and BB_{1}O are Archimedes triangles with inner triangles (filled triangles in the above applet) inside the parabola segment. The combined area of the filled triangles is half that of their Archimedes progenitors: 1/2·1/4. This is the second term of the progression. The next term comes from four smaller Archimedes triangles with the total area of 1/4 of the two preceding Archimedes triangles (which was 1/4.) The combined area of their inner triangles is therefore 1/2·(1/4)^{2}, etc. Continuing this process, the total, which is the area of the parabola segment, is

1/2 (1 + 1/4 + (1/4)^{2} + ...) = 2/3.

### References

- H. Dörrie,
*100 Great Problems Of Elementary Mathematics*, Dover Publications, NY, 1965 - S. Stein,
*Archimedes: What Did He Do Besides Cry Eureka?*, MAA, 1999

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