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 June 20th, 2012, 02:28 AM #1 Member   Joined: Jan 2012 Posts: 52 Thanks: 0 Max. area of a triangle in a circle Hi, I'm very confused by this problems involving maximum and minimum areas of plane figures. I have no clue on how to solve it: - Be A and B two points of the circumference G: x² + y² = 1. If C(k, z) belongs to G, find the k and z for the maximum and the minimum for the area of the triangle ABC in terms of A(p,m) B(c,d).
 June 20th, 2012, 08:34 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Max. area of a triangle in a circle The maximum area of the triangle will occur when point C lies on the diameter of the circle passing through the mid-point of A and B. Do you see how this maximizes the altitude of the triangle, where the base is the fixed chord AB? The equation of the line coinciding with this diameter is: $y=\frac{m+d}{p+c}x$ Substituting for y into the equation of the circle G, we find: $x=\pm\frac{p+c}{\sqrt{(p+c)^2+(m+d)^2}}$ $y=\pm\frac{m+d}{\sqrt{(p+c)^2+(m+d)^2}}$ Now, we choose the point having the greater distance from the mid-point of A and B, for which the distance formula immediately yields: $$$k,z$$=$$-\frac{p+c}{\sqrt{(p+c)^2+(m+d)^2}},-\frac{m+d}{\sqrt{(p+c)^2+(m+d)^2}}$$$ The minimum area of zero will occur for the degenerate triangles caused by C coinciding with either A or B, i.e.: $$$k,z$$=$$p,m$$$ or $$$k,z$$=$$c,d$$$
 June 20th, 2012, 09:18 AM #3 Member   Joined: Jan 2012 Posts: 52 Thanks: 0 Re: Max. area of a triangle in a circle Thanks, Mark. Being direct on the spot, the correct answer of the test below (from where I got the problem) would be letter E then? - Be ABC a triangle, with A=(0,4) and B=(2,3) and C=(m,n) any point in the circumference $x^2 + y^2= 5$. The value of m that makes the triangle's area the minimum is...? a) -1 b) -3/4 c) 1 d) 3/4 e) 2 Obs: do they put the circumference's formula just to confuse us?
 June 20th, 2012, 11:37 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Max. area of a triangle in a circle This problem is different because A and B are not on the circumference of the circle. They are both outside of the circle. If we use the fixed segment AB as the base of the triangle, then the altitude can be found from the formula for the perpendicular distance between a point and a line. The line segment AB coincides with the line: $y=-\frac{x}{2}+4$ and the point is: $$$m,\sqrt{5-m^2}$$$ This is where the equation of the circle is necessary, so we may express n as a function of m. and then the perpendicular distance between the point and the line, which will be the altitude h of the triangle, is given by: $h=\frac{\|-\frac{1}{2}m+4-\sqrt{5-m^2}\|}{\sqrt{$$-\frac{1}{2}$$^2+1}}=\frac{\|m+2\sqrt{5-m^2}-8\|}{\sqrt{5}}$ Now, to find the m which minimizes h, we may use differential calculus on the expression within the absolute value (where we assume 0 < m). I hope this problem is from a course in calculus. If it is, please confirm, and I will move the topic accordingly, If not, then we will have to look for a pre-calc technique. $\frac{d}{dm}$$m+2\sqrt{5-m^2}-8$$=1-\frac{2m}{\sqrt{5-m^2}}=\frac{\sqrt{5-m^2}-2m}{\sqrt{5-m^2}}=0$ Hence: $2m=\sqrt{5-m^2}$ $5m^2=5$ We take the positive root, as the negative root is extraneous, to find: $m=1$
 June 20th, 2012, 11:45 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Max. area of a triangle in a circle Here is a much simpler method, which does not use the calculus. I feel silly for having overlooked this before. We may observe that point C must lie on the line through the center of the circle and which is perpendicular to segment AB. This line is: $y=2x$ Now substituting this into the equation of the circle, we find: $x^2+(2x)^2=5$ $5x^2=5$ $x=\pm1$ We take the point closer to the segment AB, which gives $m=1$. The root $m=-1$ gives the maximum area.

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