1957 AHSME Problems/Problem 35
Problem
Side 
of right triangle 
is divided into 
equal parts. Seven line segments parallel to 
are drawn to 
from the points of division.
If 
, then the sum of the lengths of the seven line segments:
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<length(segment(A,B)); x += length(segment(A,B))/8) { pair[] y = intersectionpoints(parallel((0,x),line(B,C)),A--C); draw((0,x)--y[0]); } // Length Label label("$10$", B/2+C/2, S); [/asy]](http://latex.artofproblemsolving.com/e/d/1/ed13166539308b489ac405436850b64a34beb94d.png)
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 
of the way along 
, its base must be of that of the larger triangle, so it must have length 
. 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 
.
Note that this answer holds for any 
such that 
, so the aforementioned ambiguity is irrelevant.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 34 |
Followed by Problem 36 | |
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| All AHSME Problems and Solutions | ||
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