Question

A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments. Determine the sum of the squares of the lengths of the three paths.

• A. 4000 m
• B. 4800 m
• C. 5600 m
• D. 6400 m
• E. 7200 m

Hint:

- In right triangles, placing legs on axes lets you use \(a^2+b^2=c^2\) directly.
- For points that divide a segment proportionally, a parameter \(t\) makes distances easy: \(P(t)=(1-t)A+tB\).
- Symmetric sums often collapse to a multiple of \(a^2+b^2\) (here, \(\tfrac{7}{8}\,c^2\)).

Answer 1

The Correct Option is C

Step 1: Model the geometry with coordinates.
Let the right angle be at the origin \(O(0,0)\). Place the legs along axes so the other vertices are \(A(a,0)\) and \(B(0,b)\). The hypotenuse \(AB\) has length \(c=80\) m, so \(a^2+b^2=c^2=80^2=6400\).
Step 2: Parameterize points dividing the hypotenuse.
Points that divide \(AB\) into four equal parts correspond to parameters \(t=\tfrac14,\tfrac12,\tfrac34\) on the segment from \(A\) to \(B\):
\[ P(t)=\big((1-t)a,\; tb\big). \] The squared length of the path \(OP(t)\) is \[ |OP(t)|^2=(1-t)^2a^2+t^2b^2. \] Step 3: Sum the three squared lengths.
\[ \sum |OP(t)|^2 = a^2\!\left[(\tfrac34)^2+(\tfrac12)^2+(\tfrac14)^2\right] + b^2\!\left[(\tfrac14)^2+(\tfrac12)^2+(\tfrac34)^2\right] = \frac{7}{8}(a^2+b^2). \] Using \(a^2+b^2=c^2=6400\),
\[ \sum |OP(t)|^2=\frac{7}{8}\cdot 6400 = 5600. \] Therefore, the sum of the squares of the three path lengths is \(\boxed{5600}\).