Hospital in a Triangle

Similar solution with @Siwat Rakngan 's and @Parth Sankhe 's solutions, presented as follows.

The problem is equivalent to finding the circumradius $R$ of the circumcircle of $\triangle ABC$ , where the hospital is located at the circumcenter. Let the sides of $\triangle ABC$ be $a$ , $b$ , and $c$ and its area $S_\triangle$ , then the circumradius is given by:

$\begin{aligned} R & = \frac {abc}{4\color{#3D99F6}S_\triangle} & \small \color{#3D99F6} \text{By Heron's formula for area of triangle} \\ & = \frac {abc}{4\color{#3D99F6}\sqrt{s(s-a)(s-b)(s-c)}} & \small \color{#3D99F6} \text{where }s = \frac {a+b+c}2 = 15 \\ & = \frac {10\times 12 \times 8}{4\sqrt{15(5)(3)(7)}} \\ & = \frac {240}{15\sqrt 7} = \frac {16}{\sqrt 7} = \frac {16\sqrt 7}7 \end{aligned}$

Therefore, $a+b+c = 16+7+7 = \boxed{30}$ .

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References:

• Circumcircle of triangle
• Heron's formula

The hospital has to be equidistant from the three towns $A$ , $B$ and $C$ . So it lies on the circumcentre of the ∆ $ABC$ . I solved it using co-ordinate geometry. I considered $town A$ as the origin and $line AB$ as the $X axis$ . Hence, B = (8,0).

I knew that the circumcentre of a triangle is the intersection point of the perpendicular bisectors of two of its sides. So, first I found out the coordinates of $C$ . Then, I decided to find the equations of the perpendicular bisectors of sides $AC$ and $AB$ (I named them as lines $l$ and $m$ respectively). And so, on solving these two equations, I got the coordinates of the circumcentre and then using the distance formula i found the circumradius of ∆ $ABC$

I found the $cos A$ using the $cosine$ $law$ . $cos A = 9/16$ . So by solving for $tan A$ using the value of $cos A$ , I found that the slope of $line AC$ is $5√7/9$ . (The slope is equal to the tangent of the angle A.)

Hence the equation of $line AC$ , which passes through the origin is $y = 5√7/9$ .

I took the coordinates of C as $(x, 5x√7/9)$ . I solved the simultaneous equations

$x^2 + y^2 =144$ and $(x - 8)^2 + y^2 = 100$

using the assumed coordinates of $C$ to obtain the actual coordinates of $C$ .

$C = (27/4,15√7/4)$ .

Then, using the $midpoint formula$ , I found the midpoint of $side AC$ , ie. $P$ , which has the coordinates ( $27/8, 15√7/8$ ). Then I found the coordinates of $Q$ , the midpoint of $side AB$ , which are $(4,0)$ . Now I had to find slopes of the lines $l$ and $m$ to form the required equations.

Now, as the product of slopes of two perpendicular lines is equal to -1, the slope of the lines $l$ is equal to the negative reciprocal of the slope of $line AC$ , ie it is $-9/5√7$ . Therefore, the equation of the perpendicular bisector $l$ in slope point form is $y - (15√7/8) = ( -9/5√7 )( x - 27/8 )$ .

Now, I have to find the equation of the line $m$ , which is easy. The coordinates of point $Q$ are $(4,0)$ . Since it is perpendicular to the X axis (ie line $AB$ ), the equation of the line $m$ is $x = 4$ .

Now, I substitute x = 4 in the equation of the line $l$ to get $y = 12/√7$ . Hence, the coordinates of the circumcentre are $( 4, 12/√7)$ Whew!

Then using the distance formula, I found out the circumradius of the triangle which is $16√7/7$ Therefore, the answer is $16 + 7 + 7 = 30$ .

When I checked my answer with those of others, I was dumbstruck! I knew that formula for the circumradius, but it didn't just occur to me.

This is my first time with LaTex and the first solution. Please apologise me for any miscodings.

We can model the towns as vertices of a triangle. Since the hospital is equidistant from all three vertices, it will be located on the triangle's circumcenter, which is also the center of a circle that surrounds the triangle.

Therefore, the distance between the hospital and each town is the radius of the circle (circumradius).

Circumradius: $r = \frac{abc}{4{A}_{T}}$ , where $a$ , $b$ , $c$ are the side lengths of the triangle and ${A}_{T}$ is its area.

The area can be found by Heron's formula: ${A}_{T} = \sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , $c$ are the side lengths of the triangle and $s$ is the semiperimeter - $s=\frac{a+b+c}{2}$ .

Substitute in the values to get the distance of $\frac{16\sqrt{7}}{7}$ . (Note: You need to rationalize the denominator to obtain this.)

Answer: $a+b+c = 16+7+7 = \boxed{30}$

The hospital will be on the circumcentre, hence equidistant from all three vertices.

Circumradius (R) = $\frac {abc}{4S}$ , where a,b,c are the sides and S is the area.

Put in the values and the answer comes out to be $\frac {16√7}{7}$