# Prove That Area of Isoceles Triangle in an Ellipse is Maximum When Vertex On The Major Axis Lies On The Line Of Symmetry of the Triangle?

This is one of 101 classes questions whose solutions can be easily found on google, but most of the solutions assume without giving any proper line of reasoning that to maximize area (unique)vertex on the major axis must lie on the line of symmetry of the triangle?

I wanted to address this question in more rigorous manner just to satisfy my curiosity, so I started on proof writing journey with the following line of reasoning:

1. Maximize the Area Provided Vertex On Major axis Lies On line of Symmetry .(EASY)
2. Maximise the Area Provided Vertex On Major Axis doesn't Lie on line of Symmetry.

(^STUCK HERE)

1. Try to establish the inequality bewteen two

Step 1 is fairly easy with loot of (almost same)solutions on internet

Step 2 starts out easy but becomes giant monstrosity

In the given problem one unique vertex is already given $$(a,0)$$

Lets find the equation of base BC of our Isoceles Triangle with Slope $$m$$ joining vertex $$C(a,0)$$ and $$B(x_1,y_1)$$ which lies on ellipse except the line of symmetry/major axis.

Here how the Step 2 Follows in broad strokes :

1. Find the solution to other base vertex $$B(x_1,y_1)$$ in terms of slope $$m$$ of $$BC$$
2. Calculate The length of base BC.
3. Find the the solution to altitude vertex in terms of slope $$m$$ of base $$BC$$
4. Calculate the length of Altitude
5. Put the terms in Area formula($$\Delta$$) and evaluate $$\frac{d \Delta}{dm} = 0$$

Solution to the Base Vertex

$$b \sqrt{1-(\frac{x_1}{a})^2} = m(x_1-a)$$ - (i)

$$b^2(1-(\frac{x_1}{a})^2) = m^2(x_1-a)^2$$ - (ii)

Solving further:

$$\frac{x_1}{a} = \frac{(am)^2 - b^2}{(am)^2 + b^2}$$ -(iii)

Calculating the Base length

$$L^2 = (x_1 - a)^2 + (y_1-0)^2$$

$$L^2 = (\frac{y_1}{m})^2 + (y_1)^2$$ -- using (ii)

Solving further((iii) is used between the skipped steps):

$$L = \frac{2ab^2}{(am)^2 + b^2}(\sqrt{1+m^2})$$

Solution to the Altitude Vertex

1. Find Normal To the base $$BC$$ , passing through its midpoint.
2. Solve The Simultaneous Equations of Normal and ellipse

Midpoint $$(\frac{x_1+a}{2},\frac{y_1}{2})$$

$$y - y_1 = -\frac{1}{m}(x-\frac{x_1+a}{2})$$

Solving further using (iii) :

We can write the equation in compact and elegant form:

$$Q=\frac{am^2(a^2-b^2)}{(am)^2+b^2}$$

$$y = \frac{(Q-x)}{m}$$

Brace yourself for the upcoming monstrosity

Solving the simultaneous equations lead to very untenable solution to Altitude vertex A$$(x_a,y_a)$$

$$m^2y^2 = Q^2+ x^2 - 2Qx$$

$$m^2b^2(1- (\frac{x}{a})^2) = Q^2+ x^2 - 2Qx$$

$$x_a = \frac{Q \pm \frac{mb\sqrt{a^2+(mb)^2 - Q^2}}{a}}{1 + (\frac{mb}{a})^2}$$

Where Am I STUCK?

Now I don't know how to tame the solution so I can find an elegant equation for calculating length of altitude.

I've tried many ways to find some mathematical relation to reduce the solution to elegant and compact form, but to no help :(

1. Tried writing the equationin parametric/trigonometric form
2. Tried using slope-angle between lines relation using isoceles properties
3. Tried using distance/parametric form of line to calculate length of altitude

Altitude = $$(x_a - x_{midpoint})\sec{\theta}$$

$$=(x_a - x_{midpoint})\sqrt{1+(\frac{-1}{m})^2}$$

Putting it in area formula and finding the roots of first derivative leads to eye-gauging solution to m, I dare not post here

^This is the best solution I'v got :(

• I think you should explain better the original problem, for instance with a link to one of those solutions you allude to. As far as I know, an isosceles triangle with maximum area can have its vertex either at an endpoint of the major axis, or at an endpoint of the minor axis, which seems to contradict what you say. Feb 9, 2019 at 18:54
• Thanks Aretino for giving your time to this. There is already one question on MSE which exactly phrases the problem: math.stackexchange.com/questions/2390641/… But it solves the problem under the same assumption without any solid reasoning :( Feb 9, 2019 at 19:30

You can think of an ellipse with semi-major axis $$a$$ and semi-minor axis $$b$$, as a circle of radius $$a$$ which is then "squeezed" by a ratio $$b/a$$ along the direction of minor axis. Any polygon inscribed in the circle becomes then a polygon inscribed in the ellipse, and its area changes by the same ratio $$b/a$$.
But we know that the triangle of maximum area inscribed in a circle is any equilateral triangle (see here for a proof), having an area of $${3\sqrt3\over4}r^2$$. It follows that any triangle of maximum area inscribed in the ellipse described above has an area of $${3\sqrt3\over4}ab$$ and must the image of an equilateral triangle after squeezing.
Among these triangles, there are only four isosceles triangles: they are the image of those equilateral triangles in the circle of radius $$a$$ having a vertex either on the minor axis, or on the major axis, for in that case two sides ares stretched by the same amount.
• I wrote "squeeze" but you can also see that the other way and stretch a circle of radius $b$ along a direction of a factor $a/b$. That is achieved, for instance, by multiplying $x$ coordinates of all point by the same scale factor, and leaving the $y$ coordinates the same. If you take the equation of a circle and apply that transformation you get the equation of an ellipse. See here: en.wikipedia.org/wiki/Scaling_(geometry) Feb 9, 2019 at 20:34
• Thank you, I 'm onto it :). I'm also exploring the idea of finding altitude of equilateral triangle in Circle and then scaling it by factor $\frac{b}{a}$ and compare it with other approaches I've tried to see if I can find more ways to tame the complexity. Feb 9, 2019 at 20:42