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Im trying to find a way to calculate the position of the one of the base points of an isocicles triangle if I know the positions of the other two points, the angle measures, and the side lengths. It must be possible since you know everything else about the triangle except that one base point. However I've yet to find success.

Is there a way to do this?

j76goatboy
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2 Answers2

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Hint.

It is sufficient to know the position of a base point $A=(x_A,y_A)$ , the position of the vertex $C=(x_C,y_C)$ and the leght of the basis $\overline{AB}=a$.

We have that the side of the triangle is $\overline{AC}=l= \sqrt{(x_A-x_C)^2+(y_A-y_C)^2}$ and the $B$ have to be the intersection of the two circle with center $C$ and radius $l$ and of center $A$ and radius $a$, given by the solution of the system: $$ \begin{cases} (x-x_A)^2+(y-y_A)^2=a^2\\ (x-x_C)^2+(y-y_C)^2=l^2 \end{cases} $$ Note that, in general we have two solutions.

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Emilio Novati
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  • If you know the angle measures only one point of the two points will satisfy. Might need to know some very basic trig to compare the slopes of intersecting lines to find the angle. – fleablood Dec 13 '15 at 21:03
  • The angles are completely determined by $a$ and $l$. The two triangle have different positions but are equal. – Emilio Novati Dec 13 '15 at 21:25
  • My point is that as you are given the base angle then you can determine which of the two points is which and ... oh, this is geometry and they'd give the base angle as less than 180 regardless of orientation. I was thinking one of the two points will form a $\theta$ degree angle and the other a $- \theta$ angle but they wouldn't list the angles as such, would they? They's list both angles as $|\theta|$. – fleablood Dec 13 '15 at 21:33
  • After a couple of pages of solving for the two solutions to circle intersection I got $ y = \frac {l + y_c - x_a + x_c - a + y_a}{2}$ I don't think this is correct since There is only one answer. – j76goatboy Dec 15 '15 at 03:17
  • It is simpler if you solve the system after a change of coordinates. Write the equations of the two circles in a system that has center $A$ and the $x$ axis on $AC$. So the equations becomes simpler and you can find the solution in the form $B_{1,2}=(x_B,\pm y_B)$. Than perform the rotation and the translation that go back to the old reference system. – Emilio Novati Dec 15 '15 at 13:50
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To do this lets break the isosceles down the middle into two right triangles. And we know that the tip of the isosceles triangle has the point $(x_1,y_1)$ and one of the base points is $(x_2,y_2)$ and you want to find the other point $(x_3,y_3)$ in terms of the other points. Well right off the bat if you know the side lengths then it can be said that $$x_3=\left\{\begin{aligned} &x_2+L &&: x_2 < x_3\\ &x_2-L &&: x_2 > x_3 \end{aligned} \right.$$ $$y_3=y_2$$ Where the condition $x_2<x_3$ basically means if the point you're trying to find is to the right of the base point you know and $x_2>x_3$ essentially means the point is to the left and $L$ is the length of the base.

Now lets say you know the angle $\theta$ on the top and the two side lengths, $S$, but not the base length, then we must solve for $L$. In that case $$L=2S\sin\left(\frac{\theta}{2}\right)$$

Will Fisher
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  • "Well right off the bat if you know the side lengths then it can be said that " Huh? No, we can't. This assumes the side in question is parrellel to the x-axis which is a thoroughly invalid assumption. – fleablood Dec 13 '15 at 21:29