Triangle

particlejohn · #1 $Old$ 08-28-2008, 02:19 PM

Let $T$ be an acute triangle. Let $R$ and $S$ be inscribed rectangles in $T$ . Also, let $A(X)$ be the area of a polygon $X$ . Does $\frac{A(R)+A(S)}{A(T)}$ have a maximum? If so, what is it? $R$ and $S$ range over all rectangles and $T$ ranges over all triangles.

So they probably gave $T$ as an acute triangle for some reason. I am not sure why we need both $R$ and $S$ , since they are both rectangles. If you just let $R$ range over all rectangles, wouldn't that "cover" $S$ ? (e.g. we could instead consider $\frac{A(R)}{A(T)}$ )? Now $A(T)$ is the area of all the triangles inside $T$ right?

Any ideas? Use any derivative tests?

particlejohn · #2 $Old$ 08-28-2008, 03:55 PM

Or you could also consider $\frac{A(S)}{A(T)}$ right?

ticbol · #3 $Old$ 08-28-2008, 06:13 PM

Quote:

Originally Posted by particlejohn $View Post$

Let $T$ be an acute triangle. Let $R$ and $S$ be inscribed rectangles in $T$ . Also, let $A(X)$ be the area of a polygon $X$ . Does $\frac{A(R)+A(S)}{A(T)}$ have a maximum? If so, what is it? $R$ and $S$ range over all rectangles and $T$ ranges over all triangles.

So they probably gave $T$ as an acute triangle for some reason. I am not sure why we need both $R$ and $S$ , since they are both rectangles. If you just let $R$ range over all rectangles, wouldn't that "cover" $S$ ? (e.g. we could instead consider $\frac{A(R)}{A(T)}$ )? Now $A(T)$ is the area of all the triangles inside $T$ right?

Any ideas? Use any derivative tests?

Reason why the two inscribed rectangles were considered may be because they don't overlap each other....which is why even only one inscribed rectangle could be considered for the purpose of the question.

Yes, the [A(R) +A(S)] / A(T) has a maximum.
The A(T) is constant.
The A(R) and/or A(S) are variable, depending on how they are positioned inside the triangle.
If the positions of A(R) +A(S) is maximized, then the [A(R) +A(S)] /A(T) is maximized also.

particlejohn · #4 $Old$ 08-29-2008, 09:42 PM

So essentially the problem can be simplified as follows: What is the largest rectangle you can inscribe in an acute triangle?

From this we know that a maximum exists.

ticbol · #5 $Old$ 08-29-2008, 10:18 PM

Quote:

Originally Posted by particlejohn $View Post$

So essentially the problem can be simplified as follows: What is the largest rectangle you can inscribe in an acute triangle?

From this we know that a maximum exists.

Yes, if only one rectangle is considered.

If there are two rectangles, and they differ from each other, they will overlap each other. I was wrong in assuming they should not overlap each other. Otherwise one wiil not be inscribed in the triangle.