In the February 1954 issue of Mathematics Teacher, Paul C. Clifford describes an optimization problem from his trigonometry class. Of all isosceles triangles ABC with sides AB and BC of length 12, which one has the maximal area? Clifford told his class that an exact solution to the question required calculus. One student, however, proved him wrong by offering a simpler approach, described here with the benefit of the Web Sketchpad model below.

Drag vertex A, and observe that base BC remains fixed in length. Because the area of a triangle is half the product of a base and its corresponding altitude, to maximize ∆ABD’s area, we must maximize the length of its altitude from vertex A to base BC. Tap Show Altitude to reveal altitude AD. By dragging vertex A, you will see a continuum of right triangles ADB, all of which share a common hypotenuse of constant length AB.

In every triangle, the length of side AD is less than the length of the hypotenuse. Only when altitude AD is positioned to coincide directly with AB, with the triangle degenerating into a segment, does altitude AD achieve its greatest length. In that configuration, ∆ABC is an isosceles right triangle with AD = AB = 12.

Another way to think about altitude AD is to consider the path traced by vertex A as you drag it. Since side AB is of constant length, the path is a circle. Viewed in relation to the circle, altitude AD is a half-chord moving in a direction perpendicular to a diameter that contains side BC. Of all such chords, a radius has the greatest length. So as before, altitude AD attains its maximum value when it coincides with radius AB. The circle also serves as a reminder that two symmetric locations of point A—one on each side of segment BC— maximize ∆ABC’s area.

A Related Problem

The websketch below shows ∆BCD with point A in its interior. Point A connects to the triangle’s three vertices, forming segments AB, AC, and AD. These three segments are constructed to remain fixed in length when you drag any vertex (try it!). How should vertices B, C, and D be positioned to maximize the area of ∆BCD?

Altitudes helped with our first triangle area problem, so let us include one here. Tap Show Altitude to reveal altitude BE. As you drag vertex B, notice that the length of base CD remains unchanged. For this particular base, the area of ∆BCD will be largest when the length of altitude BE is maximized.

Tap Show Parallel to reveal a line parallel to CD through point A that intersects altitude BE at point F. Regardless of vertex B’s location, the length of segment EF is constant. To maximize the remainder of the altitude, BF, we must, as before, drag vertex B until segment AB aligns with BF. This action accounts for one vertex of ∆BCD but leaves the other two unattended.

Incremental Improvements

The websketch below displays all three altitudes of ∆BCD. Drag vertex B to align segment AB with the altitude through point B. Next drag vertex C so that its altitude coincides with segment AC. Repeat this process with vertex D.

Notice that whenever you “fix” one vertex, the others fall slightly out of alignment. Nonetheless, because each adjustment maximizes an altitude while keeping the corresponding base fixed in length, the process yields a steady, incremental improvement in the area of ∆BCD.

Work your way around the triangle again, dragging vertices B, C, and D into place until all three altitudes align with segments AB, AC, and AD. In this arrangement, you have found ∆BCD’s maximum area. Dragging any vertex will shorten the corresponding altitude and therefore decrease the area of the triangle.

Final Thoughts

The method for maximizing ∆BCD’s area depends on the interactive nature of the on-screen triangles. Indeed, it is hard to imagine a textbook conveying the optimization technique through static pictures alone. The interactivity on display is no substitute for a proof; it goes hand in hand with careful reasoning. Technology has allowed us to pursue a proof path that would be difficult to visualize purely in our minds. Not all problems benefit from this technological boost, of course, but I suspect that many similar examples can be found.