A geometry problem by Adrian Taboada

Bashing a little with law of cosines, we get the following formula:

$L=\sqrt{\dfrac{a^2+b^2+c^2+\sqrt{3(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{2}}$

where $L$ is the side of the equilateral triangle and $a$ , $b$ and $c$ are the distances from $P$ to the vertices of the triangle.

Substituting we get:

$L=\sqrt{\dfrac{4^2+5^2+6^2+\sqrt{3(4+5+6)(4+5-6)(4-5+6)(-4+5+6)}}{2}}$

$L=\sqrt{\dfrac{77+\sqrt{4725}}{2}}$

$L=\sqrt{\dfrac{77+15\sqrt{21}}{2}}$

$L \approx 8.5363$

Let the equilateral triangle $\Delta ABC$ be such that $AB = BC = AC = x, AP = 4, BP = 5$ and $BC = 6$ .

Now let $\angle APB = \alpha$ and $\angle APC = \beta$ . Then $\angle BPC = 2\pi - (\alpha + \beta)$ . Now use the Cosine Law on triangles $\Delta APB, \Delta APC$ and $\Delta BPC$ to establish the respective equations

(i) $x^{2} = 41 - 40\cos(\alpha)$ ,

(ii) $x^{2} = 52 - 48\cos(\beta)$ , and

(iii) $x^{2} = 61 - 60\cos(\alpha + \beta).$

This system has (second quadrant) solution values $\alpha = 2.49267, \beta = 2.02059$ radians. Plugging this value of $\alpha$ into (i) yields the solution $x = \boxed{8.536}$ to $3$ decimal places.

(Note that the same value for $x$ is found when this value for $\beta$ is plugged into (ii).)

Put the equilateral triangle $\triangle ABC$ in a Cartesian plane, with one of the vertices say C coinciding with the origin. Therefore, $A=(l/2,\sqrt{3}l/2)$ and $B = (l, 0)$ , where $l$ is the length of the side of the triangle. Let P have coordinates $(x, y)$ .

Now, compute the distances of the vertices to P. I assume $CP = 6$ , $AP = 5$ and $BP = 4$ . One has: $x^2 + y^2 = 36\;,\\ (l - x)^2 + y^2 = 16\;,\\ (x - l/2)^2 + (\sqrt{3}l/2 - y)^2 = 25\;.$

Manipulating this system of equations, one arrives at the following biquadratic equation: $l^4 - 77l^2 + 301 = 0\;.$

The only viable solution is $l = \sqrt{\frac{15 \sqrt{21}}{2}+\frac{77}{2}} \approx \boxed{8.53635}$ . The other positive solution is $l = 2.03241$ but it is unacceptable, since it would imply P to be outside the triangle.

The relation,

$3({p}^{4}+{q}^{4}+{r}^{4}+{a}^{4})={({p}^{2}+{q}^{2}+{r}^{2}+{a}^{2})}^{2}$ is always true for an equilateral triangle.

Where, $p,q,r$ are the distances of the point from the vertices, $a$ is the length of the side of the triangle.

This relation can be proved using co-ordinate geometry and some bashing.

After this we can substitute the values and get the answer as $8.5364$ .

I have applied coordinate geometry to solve this problem.

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Consider the equilateral shown above in the Cartesian plane with side length $a$ units with vertices at $(0,0)$ , $\left(\dfrac a2 \, \dfrac{\sqrt 3 a}{2}\right)$ and $(a,0)$ . Let coordinates of position of $P$ be $(x,y)$ with distances $5$ , $4$ and $6$ units respectively from the three vertices.

Using distance formula, we get,

${\left( x - \dfrac a2 \right)}^2 + {\left( y - \dfrac{\sqrt3 a}{2} \right)}^2 = 16 \ \cdots (1) \\ x^2 + y^2 = 25 \ \cdots (2) \\ {(x-a)}^2 + y^2 = 36 \ \cdots (3)$

Expanding $(1)$ , we get,

$x^2 + \dfrac{a^2}{4} - ax + y^2 + \dfrac{3 a^2}{4} - \sqrt{3} ay = 16 \ \cdots (4)$

Expanding $(3)$ , we get,

$x^2 + a^2 -2ax + y^2 = 36 \ \cdots (5)$

Substituting $x^2 + y^2$ with $25$ [from $(2)$ ] in $(4)$ and $(5)$ , we get,

$25 + a^2 - ax - \sqrt{3} ay = 16 \implies a^2 - ax - \sqrt{3} ay = -9 \ \cdots (6) \\ 25 + a^2 -2ax = 36 \implies a^2 -2ax = 11 \ \cdots (7)$

Using $(6)$ and $(7)$ , we obtain the value of $x$ and $y$ in terms of $a$ as:

$\color{#20A900}{x = \dfrac{a^2 - 11}{2a} \\ y = \dfrac{a^2 + 29}{2 \sqrt3 a}}$

Plugging in the values of $x$ and $y$ in $(2)$ , we obtain,

${\left( \dfrac{a^2 - 11}{2a} \right)}^2 + {\left( \dfrac{a^2 + 29}{2 \sqrt3 a} \right)}^2 = 25$

$\implies a^4 - 77 a^2 + 301 = 0$

Whose, positive roots (as we are required to find magnitude of length, so we are considering only positive roots) are given as:

$a = \sqrt{\dfrac{77 + 15 \sqrt{21}}{2}} \ \text{or} \ a = \sqrt{\dfrac{77 - 15 \sqrt{21}}{2}}$

The value of $\sqrt{\dfrac{77 - 15 \sqrt{21}}{2}}$ comes out to be approximately $2.03$ , which implies that $P$ lies outside our triangle as each of the distances from $P$ to each of the vertices is greater than $2.03$ .

But, since, $P$ lies within our triangle, so we have,

$a = \sqrt{\dfrac{77 + 15 \sqrt{21}}{2}} \approx \boxed{8.53}$

i had to use wolfram alpha you can see it here : " https://www.wolframalpha.com/input/?i=sqrt(x%5E2-1) sqrt(81-x%5E2)%2B+sqrt(x%5E2-4) sqrt(100-x%5E2)%2Bsqrt(x%5E2-1) sqrt(121-x%5E2)-sqrt+(3) x%5E2%3D0 " i raeched these equations using the area constraint.