<p><strong>The Quadrature of the Parabola</strong> is a treatise on geometry written by Archimedes in the third century, in which 24 propositions regarding parabolas were proven.  The last two results shown that the area of a parabolic segment (region enclosed by a parabola and a chord) is \(\frac{4}{3}\) that of a certain inscribed triangle.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1770153043323-ud5l1fn7vt8.png" style="max-width: min(200px, 100%); height: auto; display: block;" alt="Figure 4:  A parabolic segment." class="aligncenter custom-width" /></figure></div>
<p>We shall show the second result which is a geometric argument that involves the sum to infinity of a certain geometric series.</p>
<p>The main idea of the proof is the dissection of the parabolic segment into infinitely many triangles, as shown in the figure below.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1770153263844-aix9cz8khx.png" style="max-width: min(300px, 100%); height: auto; display: block;" alt="Figure 5: Archimedes' second proof dissects the area using an arbitrary number of triangles." class="aligncenter custom-width" /></figure></div>
<p>From the earlier results presented in the treatise, Archimedes proves that the area of each green triangle is \(\frac{1}{8}\) of the area of the blue triangle, so that both green triangles together sum to \(\frac{1}{4}\) the area of the blue triangle.</p>
<p>This can be seen as a result of the fact that the green triangle has&nbsp;\(\frac{1}{2}\) the width and \(\frac{1}{4}\) the height of the blue triangle.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1770153459704-3yr4yl2fz6c.png" style="max-width: min(400px, 100%); height: auto; display: block;" alt="Figure 6: Comparing the blue triangle and the green triangle." class="aligncenter custom-width" /></figure></div>
<p>Using a similar argument, each of the \(4\) yellow triangles has&nbsp;\(\frrac{1}{8}\) the area of a green triangle or \(\frac{1}{64}\) the area of the blue triangle, summing to \(\frac{4}{64} = \frac{1}{16}\) the area of the blue triangle; each of the \(2^3 = 8\) red triangles has&nbsp;\(\frac{1}{8\) the area of a yellow triangle, summing to \(\frac{2^3}{8^3} = \frac{1}{64}\) the area of the blue triangle, and so on.</p>
<p>Using the Archimedean method of exhaustion (sum to infinity), it follows that the total area of the parabolic segment is given by the sum to infinity of the following geometric series:<br>\[
T + \frac{1}{4}T + \frac{1}{4^2}T + \frac{1}{4^3}T + \cdots,
\]<br>where&nbsp;\(T\) represents the area of the large blue triangle, the second term represents the total area of the \(2\) green triangles, the third term represents the total area of the \(2^2\) yellow triangles, and so forth. </p>
<p>Applying the sum to infinity formula of a convergent geometric series (since the common ratio is \(\frac{1}{4}\) is less than \(1\)), the area of the parabolic segment is given by<br>\[
\left( \sum_{k=0}^\infty \frac{1}{4^k} \right) T = \left(\frac{1}{1-\frac{1}{4}}\right) T = \frac{4}{3}T.
\]</p>
<p style="text-align: right">\(\Box\)</p>

**The Quadrature of the Parabola** is a treatise on geometry written by Archimedes in the third century, in which 24 propositions regarding parabolas were proven.  The last two results shown that the area of a parabolic segment (region enclosed by a parabola and a chord) is \(\frac{4}{3}\) that of a certain inscribed triangle.

We shall show the second result which is a geometric argument that involves the sum to infinity of a certain geometric series.

The main idea of the proof is the dissection of the parabolic segment into infinitely many triangles, as shown in the figure below.

From the earlier results presented in the treatise, Archimedes proves that the area of each green triangle is \(\frac{1}{8}\) of the area of the blue triangle, so that both green triangles together sum to \(\frac{1}{4}\) the area of the blue triangle.

This can be seen as a result of the fact that the green triangle has \(\frac{1}{2}\) the width and \(\frac{1}{4}\) the height of the blue triangle.

Using a similar argument, each of the \(4\) yellow triangles has \(\frrac{1}{8}\) the area of a green triangle or \(\frac{1}{64}\) the area of the blue triangle, summing to \(\frac{4}{64} = \frac{1}{16}\) the area of the blue triangle; each of the \(2^3 = 8\) red triangles has \(\frac{1}{8\) the area of a yellow triangle, summing to \(\frac{2^3}{8^3} = \frac{1}{64}\) the area of the blue triangle, and so on.

Using the Archimedean method of exhaustion (sum to infinity), it follows that the total area of the parabolic segment is given by the sum to infinity of the following geometric series:
\[
T + \frac{1}{4}T + \frac{1}{4^2}T + \frac{1}{4^3}T + \cdots,
\]
where \(T\) represents the area of the large blue triangle, the second term represents the total area of the \(2\) green triangles, the third term represents the total area of the \(2^2\) yellow triangles, and so forth. 

Applying the sum to infinity formula of a convergent geometric series (since the common ratio is \(\frac{1}{4}\) is less than \(1\)), the area of the parabolic segment is given by
\[
\left( \sum_{k=0}^\infty \frac{1}{4^k} \right) T = \left(\frac{1}{1-\frac{1}{4}}\right) T = \frac{4}{3}T.
\]

Archimedes' quadrature of the parabola

Archimedes' quadrature of the parabola