# 1957 AHSME Problems/Problem 35

## Problem

Side $AC$ of right triangle $ABC$ is divided into $8$ equal parts. Seven line segments parallel to $BC$ are drawn to $AB$ from the points of division. If $BC = 10$, then the sum of the lengths of the seven line segments:

$\textbf{(A)}\ \text{cannot be found from the given information} \qquad \textbf{(B)}\ \text{is }{33}\qquad \textbf{(C)}\ \text{is }{34}\qquad\textbf{(D)}\ \text{is }{35}\qquad\textbf{(E)}\ \text{is }{45}$

## Solution

$[asy] import geometry; point B = (0,0); point A = (0,16); point C = (10,0); // Triangle ABC draw(triangle(A,B,C)); dot(A); label("A",A,NW); dot(B); label("B",B,SW); dot(C); label("C",C,SE); // Parallel Lines for (real x=0; x

Because the location of the right angle within the triangle is not specified, answer choice (A) may be tempting. However, as we will come to see later, this ambiguity is irrelevant to the problem. Because we have a series of parallel lines, we have many smaller triangles which are similar to the larger triangle. Because the smallest triangle at the top has its base $\tfrac1 8$ of the way along $\overline{AC}$, its base must be $\tfrac1 8$ of that of the larger triangle, so it must have length $\tfrac1 8 \cdot 10$. We can repeat this for each of the seven segments and thus find our desired sum (by using the formula for triangular numbers): \begin{align*} &10 \cdot \frac1 8 + 10 \cdot \frac2 8 + ... + 10 \cdot \frac7 8 \\ = &\frac{10}8(1+2+...+7) \\ = &\frac{10}8(\frac{7 \cdot 8}2) \\ = &10 \cdot \frac7 2 \\ = &35 \end{align*} Thus, our answer is $\boxed{\textbf{(D)} \text{ is } 35}$. Note that this answer holds for any $\triangle ABC$ such that $BC=10$, so the aforementioned ambiguity is irrelevant.