# Triangle of Maximum Area

Leo Giugiuc posted an interesting problem due to Edson Curahua Ortega (Peru) at the CutTheKnotMath facebook page. The solution (Solution 1) is by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc. Solution 2 has been suggested by Kunihiko Chikaya.

In $\Delta ABC,\;$ $AB=9,\;$ $AC=41k,\;$ $BC=40k,\;$ $k\gt 0.\;$ Find $k\;$ for which $[\Delta ABC]\;$ is maximum.

Note that the problem makes perfect sense. The area if $\Delta ABC\;$ can be thought of as a function of the altitude from $C.\;$ That altitude is a monotone increasing function of $k,\;$ for $\displaystyle k\gt\frac{1}{9},$ that is, where the triangle is at all defined, up to a position where $\angle BAC\;$ is around $45^{\circ}$ (the angle is a monotone decreasing function for $k\gt 1.\;$ The decrease in the angle is compensated by the growth of the side lengths of $AC\;$ and $BC,$ such that the altitude and, with it, the area, grow, but only up to a point.)

Both the altitude and the area are continuous functions of $k\;$ which insures the existence of the maximum for, say, $k\gt 1.$

Solutions

In $\Delta ABC,\;$ $AB=9,\;$ $AC=41k,\;$ $BC=40k,\;$ $k\gt 0.\;$ Find $k\;$ for which $[\Delta ABC]\;$ is maximum.

### Solution 1

We'll use Heron's formula for a triangle with side lengths $a,b,c\;$ and the area $S\;$ in the form

$16S^2=2a^2b^2+2b^2c^2+2c^2a^2 - a^4-b^4-c^4$

which, in our case ($a=9,\;$ $b=40k,\;$ $c=41k,)\;$ becomes

$16S^2=9^2(-9^2k^4+2\cdot(41^2+40^2)k^2-9^2).$

On the right, there is a biquadratic function in $k.\;$ To remind, the quadratic function $f(t)=\alpha t^2+2\beta t+\gamma\;$ achieves its maximum (if $\alpha\lt 0)\;$ for $\displaystyle t=\frac{-\beta}{\alpha}\;$ such that $16S^2$ attains its maximum for $\displaystyle k^2=\frac{41^2+40^2}{9^2}.\;$ The maximum equals

\displaystyle\begin{align} 16S^2 &= 9^2\left[ -9^2\left( \frac{41^2+40^2}{9^2} \right)^2 + 2\cdot\frac{(41^2+40^2)^2}{9^2}-9^2\right]\\ &= 9^2\left[\frac{(41^2+40^2)^2}{9^2}-9^2\right]\\ &= 9^2\left[\frac{(41^2+40^2)^2}{9^2}-\frac{(41^2-40^2)^2}{9^2}\right]\\ &= 4\cdot 41^2\cdot 40^2 \end{align}

(because $9^2=41^2-40^2.)\;$ It follows that $\max S=2\cdot 41\cdot 10=820$ is attained at $\displaystyle k=\frac{\sqrt{41^2+40^2}}{9}.$ It can be checked now that $\sin\angle BAC\approx 0.698,\;$ making the angle close to $45^{\circ}.$

### Soluion 2

For any $k,\;$ point $C\;$ lies on the circle of Apollonius defined by $AC:BC=41:40.\;$ Let point $X,$ between $A\;$ and $B,\;$ be such that $AX:XB=41:40\;$ and point $Y,\;$ to the right of $B,\;$ satisfy $AY:BY=41:40.\;$ Solving the two equations we obtain $BY=40\cdot 9.\:$ $XY$ serves as a diameter of the circle of Apollonius whose center $O$ is midway between $X$ and $Y\;$ and the radius $\displaystyle R=\frac{1}{2}\left(\frac{40}{9}+40\cdot 9\right)=\frac{40\cdot 41}{9}.\;$ This is also the maximum possible altitude for $\Delta ABC.\;$ It follows that $\displaystyle\max S=\frac{1}{2}\cdot 9\cdot\frac{40\cdot 41}{9}=820.$

### A Sample of Optimization Problems III

• Mathematicians Like to Optimize
• Mathematics in Pizzeria
• The Distance to Look Your Best
• Building a Bridge
• Linear Programming
• Residence at an Optimal Distance
• Distance Between Projections
• Huygens' Problem
• Optimization in a Crooked Trapezoid
• Greatest Difference in Arithmetic Progression
• Area Optimization in Trapezoid
• Minimum under Two Constraints
• Optimization with Many Variables
• Minimum of a Cyclic Sum with Logarithms
• A Problem with a Magical Solution from Secrets in Inequalities
• Leo Giugiuc's Optimization with Constraint
• Problem 4033 from Crux Mathematicorum
• An Unusual Problem by Leo Giugiuc
• A Cyclic Inequality With Constraint in Two Triples of Variables
• Two Problems by Kunihiko Chikaya
• An Inequality and Its Modifications
• A 2-Variable Optimization From a China Competition