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Find the triangle of smallest area in the half-plane to the right of the y-axis whose three sides are respectively segments of the $x$-axis, the diagonal $y = x$ , and a line through $(2, 1)$. [This is from MIT's Single Variable Calculus Course].

This is my attempt at the answer:

I'm making the assumption that one vertex of the triangle is at the origin. I'm not sure this is right since it says one side is a segment of the $x$-axis. Then, the points $(2,1), (a, a)$, and $(b,0)$ must lie on the line with the point $(2,1)$. And the points $(0,0), (a,a)$, and $(b,0)$ are the vertices of the triangle.

Trying to find an equation using the points, I have: $$y - a = \frac{1-a}{2-a}(x-a)$$

Then $$0 = \frac{1-a}{2-a}(x-a) + a$$ and$$ x= b = \frac{a}{a-1}$$

Using Heron's formula for the area of the triangle,

$$A = \frac{1}{2}\left|ab\right| = \frac{a^2}{2-2a}$$ assuming 2a < 2 (I found the same value for a, when I assumed 2a > 2)

$$A'= \frac{4a - 2a^2}{(2-2a)^2}$$

Setting $$A'= \frac{4a - 2a^2}{(2-2a)^2} = 0$$ a = 0 and a = 2

Plugging in a = 0 into the original formula for A gives 0. Plugging in a = 2, gives $$A = \left|\frac{2^2}{2-2(2)}\right|\ = 2$$

I would please like to know if I am on the right track or if I have made faulty assumptions.

Thank you!

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  • $\begingroup$ Your triangle has points (0,0), (2,2) and (2,0). Doesn't that give area = 2? $\endgroup$ – c.. Mar 5 '17 at 23:33
  • $\begingroup$ Yes, you're right c.z. I was multiplying the area by another factor of $$\frac{1}{2} $$. Thank you for pointing this out! $\endgroup$ – dorkichar Mar 6 '17 at 0:00
  • $\begingroup$ Everything else looks like it's correct $\endgroup$ – c.. Mar 6 '17 at 0:10
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The line of proof is right, though the proof is not complete to the last detail. Some annotations...

I'm making the assumption that one vertex of the triangle is at the origin. I'm not sure this is right since it says one side is a segment of the x-axis.

It is right because you are given that two sides lie on the $x$-axis and the diagonal $y=x$ respectively. Any two sides of a triangle have a vertex in common, and in this case that vertex can only be the intersection of the two lines, which is the origin.

Then $\;\;\cdots$ $$b = \frac{a}{a-1}$$

Note that the triangle must be "in the half-plane to the right of the y-axis", which means $a \gt 0$ and $b \gt 0$ (neither can be $0\,$, otherwise the triangle would be degenerate). Then $b=\frac{a}{a-1} \implies a \gt 1$.

Using Heron's formula for the area of the triangle,

$$A = \frac{1}{2}\left|ab\right| = \frac{a^2}{2-2a}$$ assuming 2a < 2 (I found the same value for a, when I assumed 2a > 2)

  • Heron's formula refers to something else. I don't know that the "half of base times height" formula has its own brand name.

  • Since $a\gt1$ per previous observation, the area is in fact $A=\cfrac{a^2}{2a-2}\,$.

$$A'= \frac{4a - 2a^2}{(2-2a)^2}$$

Alternative without calculus:

$$ A=\frac{1}{2}\frac{a^2-1+1}{a-1}=\frac{1}{2}\left(a+1+\frac{1}{a-1}\right)=\frac{1}{2}\left(2 + (a-1)+\frac{1}{(a-1)}\right) \ge \frac{1}{2}(2+2)=2 $$

The latter inequality follows from $x + \frac{1}{x} \ge 2$ for $\forall x \gt 0$ with equality iff $x=1$. Therefore the minimum area is $2\,$, and is attained when $a-1=1 \iff a=2\,$.

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    $\begingroup$ Thank you so much, dxiv! This gave me a lot more insight into the problem. $\endgroup$ – dorkichar Mar 6 '17 at 5:55

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