MHB Find the altitude of the triangle with minimum area

AI Thread Summary
The discussion focuses on finding the altitude of an isosceles triangle circumscribed about an ellipse when its area is minimized. By stretching the vertical axis of the ellipse, it transforms into a circle, leading to the conclusion that the triangle achieves minimum area when it is equilateral with a height of 3a. After reverting the vertical axis to its original state, the height of the minimizing triangle becomes 3b. The conversation highlights the mathematical reasoning behind these transformations and their implications for triangle dimensions. The solution emphasizes the relationship between the ellipse and the triangle's area optimization.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Consider the isosceles triangle circumscribed about the ellipse:

View attachment 1220

Find the altitude of the triangle when its area is minimized.
 

Attachments

  • ellipseintriangle.jpg
    ellipseintriangle.jpg
    6.3 KB · Views: 116
Mathematics news on Phys.org
[sp]Stretch the vertical axis by a factor $a/b$. Then the ellipse becomes a circle of radius $a$, and the area of the triangle is minimised when the triangle is equilateral with height $3a$. Now shrink the vertical axis back to where it started (by a factor $b/a$), and the minimising triangle then has height $3b.$[/sp]
 
Opalg said:
[sp]Stretch the vertical axis by a factor $a/b$. Then the ellipse becomes a circle of radius $a$, and the area of the triangle is minimised when the triangle is equilateral with height $3a$. Now shrink the vertical axis back to where it started (by a factor $b/a$), and the minimising triangle then has height $3b.$[/sp]

Extremely clever, Chris, but then I would expect no less from you! (Sun)

Here is my solution:

I began by orienting the $xy$ axes at the center of the ellipse, and considered the right half of the triangle only. I then labeled the altitude of the resulting right triangle as $h$ and the length of its base as $B$.

The equation of the ellipse is then:

(1) $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

and the hypotenuse of the right triangle lies along the line:

(2) $$y=-\frac{h}{B}x+(h-b)$$

Substituting for $y$ from (2) into (1), we obtain the quadratic in $x$:

$$\left(\frac{a^2h^2}{B^2}+b^2 \right)x^2+\left(\frac{2a^2h(b-h)}{B} \right)x+a^2h(h-2b)=0$$

Since the ellipse is tangent to the hypotenuse, we require the discriminant to be zero, which implies:

$$h=\frac{2bB^2}{B^2-a^2}$$

And so the area $A$ of the right triangle may be written as as function of $B$ as follows:

$$A(B)=\frac{bB^3}{B^2-a^2}$$

Now, differentiating with respect to $B$ and equating the result to zero, we obtain:

$$A'(B)=\frac{bB^2\left(B^2-3a^2 \right)}{\left(B^2-a^2 \right)^2}=0$$

Since we must have $$a<B$$ we find:

$$B^2=3a^2$$

and hence:

$$h=\frac{2b\left(3a^2 \right)}{3a^2-a^2}=3b$$

It is easy to see this minimizes the area of the triangle by the first derivative test.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top