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.