The Triangular Room

Geometry Level 4

Daniel is standing in the middle of an equilateral triangle-shaped room. He wants to touch all three walls and return back to the center of the room. If the shortest distance between Daniel and a wall is $1$ , then $d$ is the shortest distance Daniel has to walk to complete his task. Find $d^2$ .

Details and Assumptions

Walking to a vertex of the triangular room counts as touching both walls adjacent to the vertex.

The answer is 28.

1 solution

Maria Kozlowska
Jul 13, 2015

Triangle side can be calculated as: $\frac{h}{3} =\frac{\sqrt{3}a}{2\times3} = \frac{a}{2\sqrt{3}}=1 \Rightarrow a=2\sqrt{3}$

The shortest route will be a line reflected. When mirror images of the room are added, the route will be segment $PP1$ That segment is equal to the route through points $PUVWP$ . Any other route (except for rotated ones) will result in a broken line with length greater then the segment $PP1$ . The length of $PP1$ is hypotenuse with short sides $\frac{h}{3}=1$ and $\frac{3 a}{2}=3\sqrt{3}$

$(PP1)^2 = (3\sqrt{3})^2 + 1 = \boxed{28}$ For more info see Fermat's principle

It would be useful if you actually showed that the length of the reflected line segment is $\sqrt{28}$

Janardhanan Sivaramakrishnan - 5 years, 11 months ago

pw + wv +uv+pu = 2+2root3 Can you explain ................

D H - 4 years, 9 months ago

$PW + WV+ UV+PU=\sqrt{28}$

Maria Kozlowska - 4 years, 9 months ago

Using 2 different vectors from the center of one triangle to the centers of the next 3 and finding the distance of the resultant . $|<2,0> + <2cos(60°),2sin(60°)> + <2,0>|^{2} \\= (4+2\cdot \frac{1}{2})^2 + (2 \frac{\sqrt{3}}{2})^2 = 25 + 3 = 28$

Jerry McKenzie - 3 years, 5 months ago

The Triangular Room

The answer is 28.

1 solution

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