00:01
Okay we have a circle of radius 5 and we want to draw a circumscribed triangle that's an isosceles triangle so it's got two sides that are equal.
00:14
Those are the two semi -vertical sides and we want to find the minimum area of that triangle.
00:26
Okay so i'm defining x1 and y1 in terms of the locations of p1 and p2.
00:33
Those are where these lines are tangent to the circle.
00:37
Down here p3 is at y equals minus 5 because we know the radius is 5 and our bottom line is perpendicular to the y axis.
00:49
So it has to be like that.
00:52
The point down here in the lower right hand corner is x2 minus 5 and the point up here where our side crosses the y axis is at 0, y2.
01:06
Actually the other side crosses there too.
01:08
But i'm only going to think about the side through p2.
01:15
So the goal is this.
01:17
I'm going to set this up to do as a lagrange multiplier problem because we're looking for a minimum.
01:24
Okay so it makes sense to use a lagrange multiplier.
01:27
We have several constraints so we're going to be minimizing the area of the triangle.
01:32
So i have to write out an expression for the area and we'll also have constraints on the variables that are going to be involved.
01:43
So line 1, l1, has this equation.
01:50
So how do we know that? we also know x1 squared plus y1 squared equals 25.
01:56
So how do we know the equation of the line? so if i draw a line from the origin out to p2, its slope is y1 over x1.
02:06
Actually it's minus y1 over x1.
02:12
No it's plus.
02:19
It's plus y1 over x1.
02:21
What am i saying? because it's got positive slope.
02:24
And so the line l1, because it's perpendicular to that line because it's tangent to the circle, has the negative reciprocal slope.
02:33
So it's minus x1 over y1 is the slope of line l1.
02:39
And so we got as a function of x gives us y.
02:42
And then y2 is the point where it crosses the y -axis.
02:49
And then like i said x1 and y1 are related because they're on that circle.
02:56
If i put y equals y1, so that's our original point, has to equal this.
03:07
And so that allows me to solve for y2.
03:13
And x1 squared plus y1 squared is 25.
03:18
So y2 is 25 over y1.
03:25
Y1 and y2 are unknown but they are related.
03:30
And see x1 and y1 are also unknown but they're related because they're on the circle.
03:38
So here's our equation for y as a function of x for the straight line.
03:44
So we need to know what happens when y is minus 5 because we need to know what x2 is.
03:54
It looks like a 3 but it's really a 2.
04:05
I did start out calling it 3...