Pick a point

Three problems in Euclidean Geometry with a common denominator:

Given a square with side-length one, consider the circle inscribed on it. Let \boldsymbol{x} be a point in the circle, and let \lambda_1, \lambda_2, \lambda_3 and \lambda_4 be the distances from this point to each of the sides of the square (in any order). Prove that \lambda_1^2+\lambda_2^2+\lambda_3^2+\lambda_4^2 is constant for any point in the circle.

A quick solution is obtained by placing the center of the square at the origin of the Euclidean plane.  In this case, the coordinates of any point in the circle can be written as (\cos \theta, \sin \theta) for different values of \theta \in [0,2\pi).  The sums of the squares of the four distances to the sides of the square should be:

(1-\cos\theta)^2 + (-1-\cos\theta)^2 + (1-\sin\theta)^2 + (-1-\sin\theta)^2 = 6.

As you can see, the value does not depend on the point in the circle: it is always 6.

The following related problems present the reader with a similar challenge:

  • What will be the value of that constant if the square has an arbitrary side-length M>0?
  • A generalization: Do we have the same property in higher dimensions? For example:

    Given a cube with side-length one, consider the sphere inscribed on it. Let \boldsymbol{x} be a point in the sphere, and let \lambda_1, \dotsc, \lambda_6 be the distances from this point to each of the faces of the cube (in any order). Prove that \lambda_1^2+\dotsb+\lambda_6^2 is constant for any point in the sphere.

One more problem, a little bit trickier:

Start with a square in the plane, and pick any point in its interior. Consider the quadrilateral constructed with the centroids of the four triangles formed this way.

  • Prove that this quadrilateral is always a square.
  • What is the ratio of the area of the big square divided by the area of the smaller square?

Interesting problem: Let me present a few steps of the construction with detail, to help the reader formulate a strategy for its solution.

Start with a square \square ABCD, any point P in its interior and consider the four triangles in which the square is divided.

Each triangle has three medians. These are the segments that join each vertex, with the midpoint of the opposite leg of the triangle. In the figure we have shown the median joining P with the midpoint M_1 of the segment \overline{AB}.

The centroid of a triangle is the intersection of its three medians. In the figure, the three medians of the triangle \triangle ABP intersect at the point Q_1.

We repeat the construction of the centroids for the remaining three triangles: This gives us centroid Q_2 in triangle \triangle BCP, centroid Q_3 in triangle \triangle CDP and centroid Q_4 in triangle \triangle ADP.

Notice what happens when we draw the quadrilateral with those four centroids as vertices. The next figure shows the result for several random choices of points P inside of the square \square ABCD:

Miscellaneous

This is the tikz code that generated the random constructions above:

\def\construction{%
  \begin{tikzpicture}
    \coordinate [label=left:\textcolor{red}{$A$}] (A) at (0,0);
    \coordinate [label=right:\textcolor{red}{$B$}] (B) at (2,0);
    \coordinate [label=right:\textcolor{red}{$C$}] (C) at (2,2);
    \coordinate [label=left:\textcolor{red}{$D$}] (D) at (0,2);
    \coordinate [label=above:$P$] (P) at ($ (1,1) + 0.5*(rand,rand) $);
    \draw[orange,very thin] (P) -- (A);
    \draw[orange,very thin] (P) -- (B);
    \draw[orange,very thin] (P) -- (C);
    \draw[orange,very thin] (P) -- (D);
    \coordinate (M1) at ($ (A)!0.5!(B) $);
    \coordinate (M2) at ($ (A)!0.5!(P) $);
    \coordinate (M3) at ($ (B)!0.5!(P) $);
    \coordinate (M4) at ($ (B)!0.5!(C) $);
    \coordinate (M5) at ($ (C)!0.5!(P) $);
    \coordinate (M6) at ($ (C)!0.5!(D) $);
    \coordinate (M7) at ($ (D)!0.5!(P) $);
    \coordinate (M8) at ($ (A)!0.5!(D) $);
    \coordinate (Q1) at (intersection of P--M1 and B--M2);
    \coordinate (Q2) at (intersection of P--M4 and B--M5);
    \coordinate (Q3) at (intersection of P--M6 and C--M7);
    \coordinate (Q4) at (intersection of P--M8 and A--M7);
    \draw (A) -- (B) -- (C) -- (D) -- cycle;
    \fill[black,opacity=0.5] (Q1) circle (1pt);
    \fill[black,opacity=0.5] (Q2) circle (1pt);
    \fill[black,opacity=0.5] (Q3) circle (1pt);
    \fill[black,opacity=0.5] (Q4) circle (1pt);
    \fill[red,opacity=0.5,draw=brown] (Q1) -- (Q2) -- (Q3) -- (Q4) -- cycle;
    \fill[black,opacity=0.5] (P) circle (1pt);
  \end{tikzpicture}
}
\begin{center}
  \begin{tabular}{cccc}
    \construction & \construction & \construction & \construction \\
    \construction & \construction & \construction & \construction
  \end{tabular}
\end{center}
Advertisements
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: