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Question: "Allison and Jay depart from the same point at the same time; Allison is traveling at 10 ft/second heading north and Jay is traveling at a speed of 8 ft/second heading east; they are connected by an elastic string. The corner of a building is located 20 feet East and 30 feet North from the point of departure. When will the elastic string FIRST touch the corner of the building (hint: use a fact about similar triangles)?"

So, the starting point is (0,0) and the corner of the building is (20,30) which forms two symmetric right triangles that share a hypotenuse as a border, both comprised of two legs of length 20 and 30 and a hypotenuse of $\sqrt{1300}$.

With that being said I had a hard time relating any "facts" of similar triangle to this question, because even though I have the length of all sides of the symmetric right triangles previously mentioned, I could not relate them to the necessarily larger triangle that would intersect the point (20,30) with it's hypotenuse. I don't know anything about this larger triangle, so how could I use any "facts" about similar triangles? Side splitting theorem, angle bisector theorem, corresponding sides, corresponding angles... none of these help me because I know nothing about the larger triangle that would first intersect (20,30) with it's hypotenuse (please correct me if I'm wrong about this).

But then I realized that a right triangle with one leg of length 60 heading North and the other of length 40 heading East must intersect the point (20,30) with it's hypotenuse (this can be proven if you draw two straight lines from the point (20,30), one to the left and the other down from (20,30), which will form two identical right triangles with legs of 20 and 30 and a hypotenuse of $\sqrt{1300}$ ). And then I related the length of these legs to the speed of the person traveling them:

The leg aligned with the North axis is 60 feet long and the person traveling it (Allison) will reach that length in 6 seconds; The leg aligned with the East axis is 40 feet long and the person traveling it (Jay) will reach that length in 5 seconds. From here I split the difference of these two times to arrive at 5.5 seconds as the answer.

Does this logic make sense? The answer key says the answer is 5.5 seconds but I'm not sure if I got lucky with this answer or if they wanted me to solve it a different way (considering the hint). Any help greatly appreciated! p.s sorry I'm describing this with words! I don't know how to draw my problem with mathjax.

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Allison travels $\frac54$ times as fast as Jay. When Jay has traveled $20$ feet, Alison has traveled $25$ feet and they are at points $B$ and $C$. The line between them has slope $-\frac 54$ and you are looking for the line passing through $A$ with slope $-\frac 54$. Using the point-slope form this is $y-30=-\frac 54(x-20)$. The intercepts of this line with the axes are $(0,55)$ and $(44,0)$. Now divide by their speeds to get $5.5$ seconds.

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  • $\begingroup$ Interesting, I completely forgot about the slop point formula. But in the pre-calc book I got this question from the slope-point formula hasn't been covered yet. Does the slope-point formula have anything to do with similar triangles? $\endgroup$
    – Matthew S.
    Oct 18 '20 at 19:25
  • $\begingroup$ And is the logic that I used to solve the time rational? $\endgroup$
    – Matthew S.
    Oct 18 '20 at 19:26
  • $\begingroup$ I believe you were just lucky. It is true that a $60-40$ triangle will go through the corner of the building, but as you realized that does not have their speeds in the proper ratio. $\endgroup$ Oct 18 '20 at 19:35
  • $\begingroup$ Thank you! Can you please tell me if your answer relates to concepts of similar triangles? Because the hint in the question really makes me think they wanted me to solve it in a way that utilizes "facts" about similar triangles and I don't know enough about this to know if the slope-point formula relates to concepts of similar triangles $\endgroup$
    – Matthew S.
    Oct 18 '20 at 19:47
  • $\begingroup$ The small and large triangles in my diagram are similar because they both reflect the ratio between the speeds. You could drop perpendiculars from the building corner to the axes to make more, which might make it easier for some people. You wouldn't need as much algebra, I think. $\endgroup$ Oct 18 '20 at 20:29

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