Geometric Representation Problem

First a word on notation: I will use the notation [AB] instead of drawing a line above AB, which is difficult to do on the computer. Please "translate" accordingly.


We start by considering the base [OP] of the isosceles triangle. 

Its length m[OP] is equal to the total height m[HJ] = 956 minus m[EP] = 67 and m[NO] = 150 :   

m[OP] = m[HJ] - m[EP] - m[NO] = 956 - 67 - 150 = 739. (1)

Let us denote I' the point in the exact middle between O and P, so 

m[OI']  =  m[I'P]  =  ½ m[OP] = 739 / 2 = 369.5.

Then also m[IJ] = m[OI'] + m[ON] =  369.5 + 150 = 519.5,

and m[HI] = m[EP] + ½ m[OP] = 67 + 369.5 = 436.5.

Since the point C is in the exact middle of IO, its "height" or distance from L is

m[LC] = m[IJ] - ¼ m[OP] = 519.5 - 739/4 = 334.75.

[The entire line IO "goes down" by half the length of the base OP, so IC which is the half of the line IO "goes down" by ½ x ½ = ¼ of the whole base length.]


This gives us the length m[KC] which is the hypotenuse of the right triangle KLC,

m[KC] = √( m[KL]² + m[LC]² ) = √( 154² + 334.75² ) = 368.5.


Next, the length m[CD] is the same proportion of the entire length m[IO]  as is the length m[LM] with respect to the entire width  m[JN] = m(HE) = 800, so:  m[CD] = m[IO] x m[LM] / 800.

We have m[LM]  = m[LN] − m[MN] = 400 - 184 =216,

and [IO] is the hypotenuse of the right triangle which is the lower half od the isosceles triangle PIO,

so  m[IO] = √( 800² + 369.5² ) = 881.2, and finally,

m[CD] = 881.2 x 216 / 800 = 237.9


For the length m(AD), since triangles PIO and AID are similar, (i.e., their sides have the same proportions) we have  m(AD) : m(OP)  =  m(ID) : m(IO)  =  m(JM) : m(JN)

So, m(AD) = m(OP) x (m(JM) / m(JN)) = 739 * (616 / 800) = 569.03


In order to know where the point F is located, we will draw a vertical line (parallel to EN) through F down to the base [JN], and let's call F' the intersection point of that parallel with JN.

That gives a large right triangle KF'F that is similar to the triangle KLC which we already know.

So,  m[KF] : m[KC] = m[F'F] : m[LC],  or:  m[KF] = m[KC] x 956 / 334.75 = 368.5 x 956 / 334.75 = 1052.4

Also,  m[KF'] : m[KL] = m[F'F] : m[LC], so  m[KF'] = m[KL] x m[F'F] / m[LC] = 154 x 956 / 334.75 = 439.8

This gives us  m[FE] = m[F'N] = m[KN] - m[KF'] =  (154 + 400) - 439.8 = 114.2

With this we also have  m[GF] = 400 - 114.2 = 285.8

The segment [AF] is the hypotenuse of the right triangle AA'F where A' is the intersection point of the segment GF with a vertical line (parallel to EN) through A. So the distance of A' from G is 

m[GA'] = m[LM] = 216, and so  m[A'F] = m[GF] - m[GA'] =  285.8 - 216 = 69.8.

We also need m[AA'] = m[EP] + (184 / 800) m[I'P] = 67 + (184 / 800) x 369.5 = 151.985

(where we go again down from P by the proportion of I'P (half the base) which corresponds to the fraction m[AP] / m[IP] (the whole side), which is equal to the fraction MN to the whole width 800).


Thus finally  m[AF] = √( m[AA']² + m[A'F]² ) = √(151.985² + 69.8² ) = 167.24

Finally, for the total distance, 

d = m[KC] + m[CD] + m[DA] + m[AF] + m[FG] =  368.5 + 237.9 + 569.03 + 167.24 + 285.8 = 1628.47