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1. in the figure , the areas of traingle cef, triangle abe, triangle adf are 3,4, and 5 respectively. find the area of triangle aef

2. equialateral triangle abc has an area of square root of 3 and side of length 2. point p is an arbitrary point in the interioir of the traignle. what is the sum of the distances from p to ab, ab, and bc?

3. in the accompanying firgure , abcd is a square . suppose be=3 cm, ec=1cm and p is a point on the diagonal bd. find the smallest possible value of pe + pc in cm.

4. in triangle abc, ab=7, ac=9. on ab, point d is taken so that bd = 3. de is drawn cutting ac in e so that quadrilateral bced has 5/7 the area of trangile abc. find ce.

5.READ THIS CAREFULLY. find the ratio between the area of a square inscribed in a circle and an equilateral circumscribed about the same circle.

1 answer

Going back to your previous posts, I did read #5 carefully when you posted these before.
At the end you said "an equilateral circumscribed about the same circle"
I can only interpret that as "an equilateral triangle circumscribed about the same circle", and that is how I did the question.

I will now do #3, where ABCD is a square, and P is any point on the diagonal BD.
Using your diagram and information I placed the square on the x-y grid, with the following points
B(0,0), E(3,0) , C(4,0), D(4,4) and A(0,4)
equation of BD is y = x
So I let P be (x,x)

sum of PE + PC
= √( (x-3)^2 + x^2) + √( (x-4)^2 + x^2)
= (2x^2 -6x+9)^(1/2) + (2x^2 - 8x + 16)^(1/2)

d(sum)/dx = (1/2)(2x^2 - 6x + 9)^(-1/2) (4x-6) + (1/2)(2x^2 - 8x + 16)^(-1/2) (4x-8)
= (2x-3)/√(2x^2 - 6x + 9) + (2x-4)/√(2x^2 - 8x + 16)
= 0 for a max/min of sum

(2x - 3)/√2x^2 - 6x + 9) = (-2x +4)/√)2x^2 - 8x + 16)
square both sides to get
(4x^2 - 12x+9)/(2x^2-6x+9) = (4x^2 - 16x + 16)/(2x^2 - 8x + 16)

cross-multiplying .....
8x^4 - 56x^3 + 178x^2 - 264x + 144 = 8x^4 -56x^3+164x^2-240x+144
which surprisingly simplifies to
14x^2 - 24x = 0
7x^2 - 12x = 0
x(7x-12) = 0
x = 0 or x = 12/7

if x = 0 , sum = √9 +√16 = 7 , ( P is at origin)

if x = 12/7 ,
..... I used by calculator to get
sum = 5 (exactly)

(There is probably a much simpler way of doing it, and I think my solution is "over-kill". As soon as I saw the "smallest possible value" I automatically kicked into Calculus mode)
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