If $AB = 9$ and $AC:BC=40:41$, then find maximum area of $\triangle ABC$. (via Brilliant)

This is a question from Brilliant.org

The triangle $ABC$ has $AB = 9$ and $AC:BC = 40:41$. What is the maximum possible area of $ABC$?

For this question, I considered the equation $A=\frac 12ab\sin\theta$.

Since $\sin\theta\le 1$, then $A$ is maximised when $\sin\theta = 1$.

This meant $ABC$ was a right-angled triangle, after some working I got the answer as 180. However, it was wrong, Brilliant said it is not necessarily a right-angled triangle and then used Heron's formula to find the maximised area, 820.

I checked other posts which were similar to my question such as, How to maximize the area of a triangle, given two sides?. However, they followed the same method I did

I am interested in why a right-angled triangle would not maximise the area in this case and what is wrong with my logic?

• You are not given two sides of the triangle, but rather different data. Somehow, with base $9$ and the constraint given, you need to maximise the height $h$ of the triangle. – Mark Bennet Jun 7 '18 at 22:15
• Well, if $a,b$ would have been fixed or independent, yes, but we have an expression with two constraints. I will try a similar question. The triangle $ABC$ has $AB=9$ and $AC:BC=1$. What is the maximum possible area of $ABC$? Please solve this problem first... – dan_fulea Jun 7 '18 at 22:15
• You aren't given a triangle with two sides. You are given a triangle with one side and a proportion to the other two. – fleablood Jun 7 '18 at 22:21
• Ohh, that makes much more sense. Thank you! – Confused Bob Jun 7 '18 at 23:00

Your confusion stems from the fact that the area is maximized at $\theta=90$, if you keep the side lengths $BC,AC$ fixed. However in this problem $BC,AC$ can vary in length.

• AB cannot vary. – Szeto Jun 7 '18 at 22:19
• @Szeto: whoops wrong side. Fixed. – Alex R. Jun 7 '18 at 22:19

Let the sides $AC=40x, BC=41x$. Using the Heron's formula: $$S=\sqrt{\frac{81x+9}{2}\cdot \frac{81x-9}{2}\cdot \frac{9+x}{2}\cdot \frac{9-x}{2}}=\frac{81}{4}\sqrt{\left(x^2-\frac 1{81}\right)\left(81-x^2\right)}\overbrace{\le}^{GM-AM} \\ \frac{81}{4}\cdot \frac{\left(x^2-\frac 1{81}\right)+\left(81-x^2\right)}{2}=820,$$ equality occurs when $x^2-\frac1{81}=81-x^2 \Rightarrow x\approx 3$.

Note: $GM-AM$ is the Geometric Mean-Arithmetic Mean inequality.

As mentioned in a comment, the best way to realize where is the error is to consider the similar problem where $AB=9$ is given and fixed, and where $AC:BC=1$. Then in this case the triangle is isosceles, and we have the option to choose its height as big as we want. The bigger this heigth $h$, (the smaller the angle in $C$, and) the bigger the area. There is no limit for the area, we can let $h$ tend to infinity, the area goes to infinity!

As seen also in this example, in the given situation the sides $a=BC$, $b=AC$ and the angle in $C$ variate together in a "complicated way", and contribute more or less to the area. A maximum for $\sin \hat C$ does not lead to a maximal area, since the sides $a,b$, and the height $h$ are dramatically constrained to be "small".

Now let us also give a simple solution to the "brilliant" problem.

First of all, let us consider the following example of a triangle satisfying the given conditions, the one with sides $9,40,41$. Yes, it has a right angle, but not there where i supposed first that the question supposed it to be... In a (bad) picture:

(Sorry, i cannot get it smaller. But maybe it is psychologically good so.)

A possible choice of $C$ is as above, in the position which makes the angle in $B$ a right angle. An other position is the symmetrical position, which makes the angle in $A$ a right angle. With the one or the other choice we get a first weak attempt to maximize the area, it is $\frac 12\cdot 9\cdot 40=180$.

Now let us try to get a better guess. Consider a point on $AB$ which divides $AB$ in two parts, $s$ and $9-s$.

In fact we allow $$s\in\Bbb R\ ,$$ for a negative value of $s$ we consider the point on $AB$ to be on the half-line $(AB$ "after" $B$.

We draw a long perpendicular in this point to $AB$, choose on it that one point (in the "upper" half-plane) $C''$, so that the sides $AC$ and $BC$ are in the given proportion: $$\frac{AC''}{BC''}=\frac{41}{40}\ .$$ This is maybe possible, maybe impossible. (For instance, for $s=9/2$ this is impossible.) If it is possible, we want to get the height $h=h(s)$ as a function of $s$, that we will maximize.

The above figure is maybe wrong, maybe $C''$ is so that the corresponding height is $\le 40$, and maybe it is there where $?C''$ is placed, well, we have to compute. (Yes, it is wrong.)

Note: This is somehow related to the posted question. It turns out, that the maximal height is obtained for a negative $s$, (which is relatively big in absolute value,) the point on the line $AB$ which realizes the maximal height is not on the segment $AB$.

Computations. Note that from $$\frac{41}{40} =\frac{AC''}{BC''}= \frac {\sin \hat B} {\sin \hat A} = \frac {h/\sqrt{h^2+s^2}} {h/\sqrt{h^2+(9-s)^2}} = \frac {\sqrt{h^2+(9-s)^2}} {\sqrt{h^2+s^2}}$$ we get a simple relation determining $h$, $$\frac{41^2}{40^2} = \frac {h^2+(9-s)^2} {h^2+s^2} \ ,$$ so \begin{aligned} \underbrace{(41^2-40^2)}_{=9^2}h^2 &= 40^2(9-s)^2-41^2s^2 \\ &= 9(40-9s)(s+360)\ . \end{aligned} The roots of the last expression are $40/9$ and $-360$. To have a positive quantity, we need $40-9s\ge 0$, so $s\le 40/9$.

The maximal value of the expression is taken exactly in the middle of the segment determined by the roots, which is $s^*=-1600/9$. For this value of the $s$-variable we get a corresponding maximal $h^*$, which is $$h^* =\frac 19\sqrt{9\cdot (40+1600)\left(-\frac {1600}9+360\right)} = \frac {1640}9\ .$$ So the maximal area is $\frac 12 \cdot AB\cdot h^*=\frac{1640}2=820$.

Since we have expressions for the side lengths as $a,b=px,c=qx$ ($a=9,p=40,q=41$), we can ignore all the angles and use Heron’s formula for the area \begin{align} S&=\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} \tag{1}\label{1} ,\\ S(x)&=\tfrac14\sqrt{2x^2a^2(q^2+p^2)-x^4(p-q)^2(q+p)^2-a^4} \tag{2}\label{2} . \end{align}

The $x$ that maximizes the area $S(x)$, also maximizes $S(x)^2$, so we can consider \begin{align} S_2(x)&=S(x)^2=\tfrac1{16}(2x^2a^2(q^2+p^2)-x^4(p-q)^2(q+p)^2-a^4) ,\\ S_2'(x)&= \tfrac14x(a^2(q^2+p^2)-(p-q)^2(q+p)^2x^2) \end{align}

And the sought maximum can be found at

\begin{align} x_{\max}&= \frac{a\sqrt{q^2+p^2}}{q^2-p^2} =\tfrac19\sqrt{3281} \approx 6.36 ,\\ S_{\max}&= S(x_{\max})=820 . \end{align}