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I want to find the x and y of the vertex P2 in below right angle triangle:

enter image description here

I have tried Thales's theorem and triangulation but they all return a very complicated formula to calculate x and y with multiple answers. Since all the parameters of the triangle are known except the vertex P2, I believe there should be only one answer to this question.

Please take into account that the triangle can be anywhere. So assumptions won't work. I saw some people were assuming one vertex on (0,0) position.

Aboriginal
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3 Answers3

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Note that $(x,y)$ can be any point on the circle with $P_1R$ as the diameter, i.e.

$$\left(x-\frac{x_{P_1}+x_R}2\right)^2+\left(y-\frac{y_{P_1}+y_R}2\right)^2 =\frac14(x_{P_1}-x_R)^2+\frac14(y_{P_1}-y_R)^2$$

Edit:

If the side lengths are known, then

$$x= x_R + d \cos\theta, \>\>\>\>\> y = y_R - d \sin\theta$$

where

$$\theta = \tan^{-1} \frac {y_{P_1}-y_R}{x_{P_1}-x_R}+ \tan^{-1} \frac hd $$

Quanto
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  • Except the two endpoints of the diameter (i.e. $P_1$ and $R$), of course. – Geoffrey Trang Dec 02 '19 at 05:39
  • Thank you for the warm welcome and the answer to my question. As I mentioned there must be only one specific x and y for the question as the sides of the triangle are known. – Aboriginal Dec 02 '19 at 05:58
  • @Farid Farzin - added the specific result to the answer given the side lengths. – Quanto Dec 02 '19 at 06:42
  • You are a genius! It works well, except I had to update x to x=xR-dcosθ – Aboriginal Dec 02 '19 at 07:06
  • @Quanto Where can I get more info about the solution? It works on certain circumstances, not for cases. I need to investigate more on the parameters and return values. – Aboriginal Dec 02 '19 at 07:33
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If you define distance between points $(P,Q)$ as $d_{PQ}$ forming the hypotenuse/circle diameter, we have the circle of Thales thm with diameter of $ d_{PP_1}$. The center of circle has coordinates

$$ h=(x_{P}+ x_{P1})/2, k=(y_{P}+ y_{P1})/2$$

Locus is the circle with equation:

$$ (x- h)^2+(y- k)^2= (d_{PQ}/2)^2 $$

Narasimham
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To avoid cumbersome subscripts, let's instead denote the coordinates of vertices $P_1$, $R$, and $P_2$ by $(a,b)$, $(c,d)$, and $(e,f)$ respectively.

For $\Delta P_1P_2R$ to be a right triangle with $P_1R$ as the hypotenuse, segments $P_1P_2$ and $P_2R$ must be perpendicular to each other. This means that their slopes must be negative reciprocals of each other, i.e. $\big(\frac{f-b}{e-a}\big)\big(\frac{f-d}{e-c}\big)=-1$. Multiplying the numerators gives $f^2-df-bf+bd$. Likewise, multiplying the denominators gives $e^2-ce-ae+ac$. Hence, $e^2-ce-ae+ac+f^2-df-bf+bd=0$. Moving the $ac$ and $bd$ terms to the right, we get $e^2-ce-ae+f^2-df-bf=-ac-bd$. Combining like terms on the left, we get $e^2-(a+c)e+f^2-(b+d)f=-ac-bd$. Finally, by completing the square, we get $\big(e-\frac{a+c}{2}\big)^2+\big(f-\frac{b+d}{2}\big)^2=\big(\frac{a+c}{2}\big)^2+\big(\frac{b+d}{2}\big)^2-ac-bd$. This is the equation of a circle with center $\big(\frac{a+c}{2},\frac{b+d}{2}\big)$ and radius $\sqrt{\big(\frac{a+c}{2}\big)^2+\big(\frac{b+d}{2}\big)^2-ac-bd}$. Any point on that circle other than $P_1$ and $R$ could be vertex $P_2$.

Geoffrey Trang
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