Figure 1

I am trying to solve the problem above by co-ordinate geometry. Here is what I have done: I flipped the figure (just for simplicity) such that the point B is to the right of AC. Then I took C as the origin and A to be the point $(0,a)$. Hence, the point B is $(2a,0)$. Also, I took the point P to be $(x,y)$. By the distance formula, I got 3 equations, which I am not able to solve:

  1. $x^2 + y^2 = 10^2 = 100$

  2. $x^2 + (y-a)^2 = 15^2 = 225$

  3. $(x-2a)^2 + y^2 = 20^2 = 400$

By substituting (1) in (2) and (3):

  1. $a^2 - 2ay = 125$

  2. $a^2 - ax = 75$.

Can this be extended further to get the values of a,x, and y? If yes, how? Also, can we solve it using pure geometry?

  • 2
    $\begingroup$ I've just edited this for formatting, as well as to embed the picture so clicking a link is no longer needed. Some basic resources on formatting posts here: Normal formatting and mathematical formatting. Of particular note: simply hitting new line isn't enough to get a line break in the output; you need two spaces before the line break, or two line breaks. $\endgroup$ – jmerry Mar 7 at 4:40

As you state, there are $3$ equations to use

$$x^2 + y^2 = 10^2 = 100 \tag{1}\label{eq1}$$ $$x^2 + (y-a)^2 = 15^2 = 225 \tag{2}\label{eq2}$$ $$(x-2a)^2 + y^2 = 20^2 = 400 \tag{3}\label{eq3}$$

You've made a good start by using substitution. You've determined that

$$a^2 - 2ay = 125 \tag{4}\label{eq4}$$ $$a^2 - ax = 75 \tag{5}\label{eq5}$$

Next, \eqref{eq4} - \eqref{eq5} gives

$$-2ay + ax = 50 \; \Rightarrow \; a\left(x - 2y\right) = 50 \; \Rightarrow \; x = \frac{50}{a} + 2y \tag{6}\label{eq6}$$

Next, note that \eqref{eq2} - \eqref{eq1} gives

$$-2ay + a^2 = 125 \; \Rightarrow \; y = \frac{a^2 - 125}{2a} \tag{7}\label{eq7}$$

Substitute this into \eqref{eq6} to get

$$x = \frac{50}{a} + 2\left(\frac{a^2 - 125}{2a}\right) \; \Rightarrow \; x = \frac{a^2 - 75}{a} \tag{8}\label{eq8}$$

You can substitute \eqref{eq7} and \eqref{eq8} into \eqref{eq1} to get a quadratic equation in $a^2$ that you can easily solve using the quadratic formula. With the value(s) of $a^2$, you can determine $a$ (as $a \gt 0$, but only use the one value that makes sense in the diagram), then get $x$ and $y$ from \eqref{eq8} and \eqref{eq7}, respectively. I trust you can finish the rest by yourself.

As for solving it using "pure" geometry, I believe you can at least use geometry along with some algebra. The way I would approach this is to draw lines from P that are perpendicular to each triangle side, i.e., AB, AC and BC. Next, I would use trigonometric functions for each smaller triangle to relate the side lengths and the angles, e.g., $\sin\left(\angle PCB\right) = \frac{y}{10}$. Also, use that certain angles are related to each other, e.g., in radians, $\angle ABC + \angle ACB = \frac{\pi}{2}$ and $\angle ACP + \angle PCB = \frac{\pi}{2}$. In addition, there are certain trigonometric identities involving things like the sum of $2$ sines being a product of trigonometric values, such as described in Product-to-sum and sum-to-product identities. This will allow you to create various equations you can try to solve, but I suspect it'll be messier and involve more work than using algebra with a co-ordinate system, like done here. However, note there might be some theorem or simpler way than what I'm proposing here.


Starting with $$x^2 + y^2 - 100 =0\tag{1}$$ $$x^2 + (y-a)^2 - 225=0 \tag{2}$$ $$(x-2a)^2 + y^2 - 400=0 \tag{3}$$ $$(2)-(1) \implies a^2-2 a y-125=0\implies y=\frac{a^2-125}{2 a}$$ $$(3)-(1) \implies 4 a^2-4 a x-300=0 \implies x=\frac{a^2-75}{a}$$

Plug in $(1)$ and simplify to get $$5 a^4-1250 a^2+38125=0$$ which is a quadratic equation in $a^2$.

  • $\begingroup$ The value of 'a' I am getting is sq. roots of (125+40√5) and (125-40√5)... there are 2 values for a $\endgroup$ – Saee Mar 7 at 6:11
  • $\begingroup$ @Saee. These are the values for $a^{2}$, not for $a$ $\endgroup$ – Claude Leibovici Mar 7 at 6:15
  • 2
    $\begingroup$ @Saee As I state in my answer, you need to be careful of which value of $a$ you get that makes sense. You'll see that only $1$ does in terms of determining positive values for $x$ and $y$. $\endgroup$ – John Omielan Mar 7 at 6:19

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