Neighboring Friends

It is clear that the original three friends' houses are represented by a triangle with side lengths $30,50,70$ . Cyril must be located at the circumcenter of this triangle. Using the formula that $[ABC] = abc/4R$ , where a, b, c are the side lengths and ABC is the area of the triangle, and Heron's formula, we can arrive at the fact that $R = \frac{70 \sqrt{3}}{3}$ , which is approximately 40.

Let Jessica's house be point $A$ , Blesilda's house be point $B$ and Jeremiah's house be point $C$ . We are effectively trying to find the radius of the circumcircle.

By Law of Cosines, $70^2=30^2+50^2-2(30)(50)\cos{\angle BAC}$ . Simplifying, we find that $\cos{\angle BAC}=-\dfrac{1}{2}$ so $\angle BAC=120^{\circ}$ .

By the extended Law of Sines, $\dfrac{BC}{\sin{\angle BAC}}=2R$ , where $R$ is the circumradius. Plugging in $BC=70$ and $\sin{\angle BAC}=\dfrac{\sqrt{3}}{2}$ , we find that $2R=\dfrac{70}{\frac{\sqrt{3}}{2}}$ , or $R=\dfrac{70}{\sqrt{3}}$ . Using a calculator, we find that $R\approx \boxed{40}$ .

I used my geometry box ...

Because Cyril's house have an equal distances to each house. So, it means that we need to find the radius of circumcircle.

We use the formula:

R= $abc/sqrt((a+b+c)(a+b-c)(a+c-b)(b+c-a))$

Plug in all the values, and we get:

R= $40.41 m$

Let $B$ stands for Blesilda's house, $J_1$ for Jessica's house, $J_2$ for Jeremiah's house, and $C$ for Cyril's house. Restating the problem, $BJ_1 J_2$ is a triangle with $BJ_1=30$ , $J_1 J_2=50$ and $J_2 B=70$ . Using cosine law, we can actually solve for angle $J_1$ . Thus, $(J_2 B)^2=(BJ_1)^2+(J_1 J_2)^2-2(BJ_1)(J_1 J_2) \cos J_1$ $70^2=30^2+50^2-2(30)(50) \cos J_1$ $4900=3400-3000 \cos J_1$ $\cos J_1=-\frac{1}{2}$ $J_1= 120^{\circ}$ Furthermore, it is given that $CB=CJ_1=CJ_2$ . If we think of $C$ as the center of a certain circle, and $CB$ , $CJ_1$ and $CJ_2$ to be the circle's radii, it turns out that this circle actually circumscribes the triangle. Thus, we can use the extended sine rule to find $CB$ . Let $a=J_2 B, A=J_1, R=CB$ . Then $\frac{a}{\sin A}=2R$ $\frac{J_2 B}{\sin J_1}=2CB$ $\frac{70}{\sin {120^{\circ}}}=2CB$ $CB=\frac{70}{2 \sin 120^{\circ}}=\frac{35}{\frac{\sqrt{3}}{2}}=\frac{70 \sqrt{3}}{3} \approx 40.4145 \approx 40$ Therefore, Cyril house, is about 40 meters from Blesilda's house.

[Sorry for not having any diagram, because I don't know yet how to put pictures here.]

obviously,the three live in houses whose outline is a triangle. circumcentre is the point at equal distance from all the 3 vertices(houses) so circumradius is the distance.

if the sides of the triangle 30,50 and 70 in order are a,b,c then the circumradius is (a b c)/(4*A) A=area of the triangle.

a = 70m , A = 120 (Cos A = (50^2 + 30^2 - 70^2)/2x50x30)

b = 50m , B = 38.21

c = 30m , C = 21.79

from triangle sine law,

a/sin A = b/sin B = c/ sin C = 2R (R = circumradius)

70/sin 120 = 2 R

R = 40.4m (40 m).

[thanks]