# A line moving along the hypotenuse of a right triangle

ABC is a triangle with sides $AB = 6 m$, $BC = 8m$, and $AC = 10m$. A line $k$ in the plane of the triangle $ABC$ moves along the segment $AC$ at the rate of $1cm$ per sec. The line starts at A and ends at $C$ and is always perpendicular to $AC$.

1. How long does it take the line to reach the point B?
2. How long does it take the line to bisect the area of the triangle?
3. What is the area of the region that the line sweeps in its movement after $6$ min and $45$ sec?
4. What is the area of the region that the line sweeps at any give moment from the start of its movement?

I have noticed that $(6,8,10)$ are Pythagorean triple and have solved the first question (6 mints is the answer), but the last three seems a bit tedious, could somebody tell me which could be the most easiest approach to solve these three?

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I guess , you should use similarity of the triangles... – pedja Feb 1 '12 at 9:33

Let $D_1$ be the point on $AC$ such that $D_1B \perp AC$. You have that by similar triangles, $AD_1 = 3.6m$, so you are correct that part 1 is 360 secs.

Observe that the area $ABD_1 = \frac12 \times 3.6 \times 4.8 m^2$, and $BCD_1 = \frac12 \times 4.8 \times 6.4 m^2$. Let $D_2$ be the point on $AC$ such that the perpendicular line through $D_2$ to $AC$ bisects the triangle in area, notice that $D_2$ is between $C$ and $D_1$. The area of the right triangle cut out by $D_2$ and the corner $C$ is going to be $\frac12 (CD_2)^2 \times \frac{6}{8}$ using similar triangles. You set this to equal to half the total area $\frac12 \times 48$ and you get $$(CD_2)^2 = 24 \times \frac{8}{6} = 32$$ so $CD_2 = 4\sqrt{2}$. So the time to get there is $(10 - 4\sqrt{2})\times 100$ seconds which is roughly 435 secs.

Question 3 you already solved as part of Question 1.

Question 4 you have to treat the times before and after passing through $D_1$ separately. Before passing through $D_1$, you get a similar triangle to a $(6,8,10)$ triangle with the $6$ side replaced by $AD =$ the time passed in seconds multiplied by 1 centimeter per second. So the area is $$\frac12 \times \frac{8}{6} \times \left(\frac{t}{100}\right)^2$$ in meter squared.

After passing through point $D_1$, the area swept is the total area minus the area remaining. This time you compute the area of a $(6,8,10)$ triangle with the $8$ side replaced by $1000 - AD$ centimeters, where $AD$ is equal to the time elapsed in seconds.

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Here is a hint. (Hint as in: it's quite a lot of work to find the precise answer, but this may help).

I claim $\triangle ABC \sim \triangle AYX$: note that the angles $\hat A$ are the same and the angles $A\hat BC$ and $A\hat YX$ are both $90^\circ$. So, if $XY$ is the sweeping line, then the area of $\triangle ABC$ is equal to that of $\triangle AYX$ multiplied with the similarity-factor, squared. In particular

$$A(\triangle ABC) \frac{ |AY|^2}{|AB|^2} = A(\triangle AYX)$$

(This allows you to determine the area of the sweeped region, before the line $XY$ passes the point $B$. After $B$ is passed, one needs a similar consideration in the other half of the triangle to have a complete description.)

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