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Sorry for the unclear title. It was difficult to explain the problem in a concise way in 150 characters.

A right triangle has the legs 3 and 4 units, respectively. Find the probability that a line segment, drawn at random parallel to the hypotenuse and contained entirely in the triangle, will divide the triangle so that the area between the line and the vertex opposite the hypotenuse will equal at least half the area of the triangle. Make a sketch of the triangle.

I am completely stuck on this question. First of all I am assuming that the legs of length 3 and 4 are the ones containing the right angle but I am not sure if this is even right. If this is the case then a line segment drawn at random will divide the area of the triangle if it intersects the two legs (call them $a$ and $b$) at exactly $\frac{1}{\sqrt{2}}$ of their length. This way both the height and width of the smaller triangle will be divided by $\frac{1}{\sqrt{2}}$ and thus the new area will be $\frac{ab}{2}$ which is half of $ab$, the original area. How do I extend this idea to answer the question fully? Is the probability just $1-\frac{1}{\sqrt{2}}$? Thanks in advance!

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    $\begingroup$ The problem is badly posed. What does it mean to draw a line segment "at random"? Probably the question setter means that the distance of the line segment from the hypotenuse is uniformly distributed between 0 and its maximum possible value. But it should have been made clear. (You are right about the legs containing the right angle.) $\endgroup$ – TonyK May 16 '13 at 21:00
  • $\begingroup$ Hey thanks for the comment, I realize it is confusing. However, I just copy pasted the exact question I got to make sure anyone who reads it knows exactly what I do $\endgroup$ – Slugger May 16 '13 at 21:04
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Having drawn a line parallel to the hypotenuse, the smaller triangle will be proportional to the larger triangle. We will assume that 'at random' means the value of the proportionality constant is chosen uniformly between $0$ and $1$. As you've noted, if it is anything larger than $\frac{\sqrt2}{2}$, the area will be at least half of the original area. Hence, the desired probability is indeed $1-\frac{\sqrt{2}}{2}$.

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  • $\begingroup$ "The value of the proportionality constant"? What does that mean? $\endgroup$ – TonyK May 16 '13 at 21:37
  • $\begingroup$ @TonyK: The ratio of a side length of the smaller triangle to the corresponding side length of the original triangle. $\endgroup$ – Jared May 16 '13 at 21:41

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