Set up the variables needed
x = AP, y = PM, z = CP, p = AC, q = CM
Based on that, we know x+y = AM
Since MP-AP = 1, we know that y-x = 1 which leads to y = x+1
We will use y = x+1 in a few substitution steps.
The ratio below can help us isolate for p
AC:CM=3:4
AC/CM = 3/4
p/q = 3/4
4p = 3q
p = 3q/4 ... we'll use this in a substitution step
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Because we have three similar triangles, we can set up the following ratio and use it to help isolate z^2
(AP)/(PC) = (PC)/(PM)
x/z = z/y
xy = z^2 ... cross multiply
x(x+1) = z^2 ... plug in y = x+1
x^2+x = z^2
z^2 = x^2+x ... we'll use this later in a substitution step
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Focus on triangle ACM. Use pythagorean theorem. Isolate q^2
(AC)^2 + (CM)^2 = (AM)^2
(AC)^2 + (CM)^2 = (AP+PM)^2
(p)^2 + (q)^2 = (x+y)^2
(p)^2 + (q)^2 = (x+x+1)^2 ... plug in y = x+1
(p)^2 + (q)^2 = (2x+1)^2
(3q/4)^2 + (q)^2 = (2x+1)^2 ... plug in p = 3q/4
(9q^2)/16 + (q)^2 = (2x+1)^2
(9q^2)/16 + (16q^2)/16 = (2x+1)^2
(9q^2+16q^2)/16 = (2x+1)^2
(25q^2)/16 = (2x+1)^2
25q^2 = 16(2x+1)^2
q^2 = (16(2x+1)^2)/25
q^2 = (16/25)*(2x+1)^2
q^2 = 0.64(2x+1)^2 .... this will be used later
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Move onto triangle APC. Use pythagorean theorem
(AP)^2 + (PC)^2 = (AC)^2
x^2 + z^2 = p^2
x^2 + z^2 = (3q/4)^2 ... plug in p = 3q/4
x^2 + (x^2+x) = (3q/4)^2 ... plug in z^2 = x^2+x
2x^2 + x = (3q/4)^2
2x^2 + x = (9q^2)/(16)
2x^2 + x = (9*0.64(2x+1)^2)/(16) ... plug in q^2 = 0.64(2x+1)^2
2x^2 + x = (5.76(2x+1)^2)/(16)
2x^2 + x = (5.76/16)(2x+1)^2
2x^2 + x = 0.36(2x+1)^2
2x^2 + x = 0.36(4x^2+4x+1)
2x^2 + x = 1.44x^2+1.44x+0.36
2x^2 + x-1.44x^2-1.44x-0.36 = 0
0.56x^2 - 0.44x - 0.36 = 0
(56/100)x^2 - (44/100)x - (36/100) = 0
note how we have an equation with only one variable now
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Use the quadratic formula to get this solution set {x = -1/2 = -0.5, x = 9/7 = 1.285714285714}
We will ignore x = -0.5 as a negative length makes no sense
Therefore, x = AP is exactly 9/7 units long
So,
y = x+1
y = 9/7+1
y = (9/7)+(7/7)
y = (9+7)/7
y = 16/7
and therefore
AM = x+y
AM = 9/7+16/7
AM = (9+16)/7
AM = 25/7
AM = 3.57142857142858
The exact length of AM is the fraction 25/7
The approximate length of AM is roughly 3.57142857142858 units.
