1

Given a convex quadrilateral $ABCD$ circumscribed about a circle of diameter $1$. Inside $ABCD$, there is a point $M$ such that $|MA|^2 + |MB|^2+ |MC|^2 + |MD|^2= 2.$ Find the area of $ABCD$.

I've tried to solve this problem but I'm unable to do as I don't understand the use of the information of the distance of the point and how to use it, any help would be really appreciated!

answer=1

my process-

i couldn't start with this sum

Pratyush
  • 41
  • 2
    One way is to simply guess a configuration which works by trying the simplest one possible: pick ABCD to be a square and see if it contains a point giving the desired sum. (I can confirm that it does.) What I don't know how to do is whether other quadrilaterals exist with this property; I haven't found any, but that's hardly dispositive. – Semiclassical Mar 08 '23 at 06:56
  • I thought of doing it that way but I wanted a general solution which would work for all such quadrilaterals. – Pratyush Mar 08 '23 at 07:07
  • 1
    I think this could be solved using vectors. (I'm busy now, I'll try myself later.) – MyMolecules Mar 08 '23 at 07:52
  • Welcome to math.SE! Here's a tutorial and reference for typesetting math on this site. – joriki Mar 08 '23 at 07:52
  • Please include in the post what you have tried. – ACB Mar 08 '23 at 08:29
  • The condition on the sum of squares should involve a circle whose center is the center of gravity of vertices A,B,C,D ; see the answer of mine here – Jean Marie Mar 08 '23 at 09:00
  • Circumscribed quadrilaterals have a certain number of properties, for example this one – Jean Marie Mar 08 '23 at 09:07
  • I couldn't make any noticeable progress, i used a+c=b+d but it didn't get me anywhere – Pratyush Mar 08 '23 at 10:18
  • I tried using your method Jean Marie but im still unable to do the entire sum – Pratyush Mar 08 '23 at 10:53
  • If this have a unique solution, then the area has to be $1$ because a square of side $1$ with $M$ at center of circle satisfy the given condition. – achille hui Mar 08 '23 at 15:22
  • One way to generalize this would be: Given such a quadrilateral $ABCD$ (known as a tangential quadrilateral in English), find the point $M$ which minimizes $|MA|^2+|MB|^2+|MC|^2+|MD|^2$. Along the way, one would hopefully show that this sum cannot be smaller than $2$ and that this is only achieved when $ABCD$ is square. – Semiclassical Mar 08 '23 at 16:12

2 Answers2

2

Rename $A,B,C,D$ as $P_1,P_2,P_3,P_4$ and extend indexing of everything by periodicity (i.e $P_0 = P_4, P_1 = P_5$ and so on).

Let $\ell_i = |MP_i|$. We are given $M$ is inside $ABCD$. So $$\begin{align}\verb/Area/(ABCD) &= \sum_{i=1}^4\verb/Area/(MP_iP_{i+1}) \\ &\le \frac12 \sum_{i=1}^4 \ell_i \ell_{i+1} \le \frac14 \sum_{i=1}^4 (\ell_i^2 + \ell_{i+1}^2)\\ &= \frac12 \sum_{i=1}^4 \ell_i^2 = 1\end{align}$$ One the other direction, we know among all quadrilaterals circumscribing a circle, the square has the smallest area${}^{\color{blue}{[1]}}$. In this case, the diameter of the circle is $1$ and the square circumscribing it has area $1$. This means

$$\verb/Area/(ABCD) \ge 1$$

Combine these, we have $\verb/Area/(ABCD) = 1$.

Notes

  • $\color{blue}{[1]}$ - For a proof, see this answer.
achille hui
  • 122,701
1

As you asked about explicitation of the method, here it is.

It isn't still a solution, but it connects point $M$ to the centroid $G$ of the vertices.

Let us rename vertices $A,B,C,D$ as $P_k$ for $k=1\cdots 4$.

Let

$$G=\frac14 \sum_{k=1}^4(P_k)$$

be their centroid.

Let us write the given constraint under a vectorial form by expanding it using dot product :

$$2=\sum_{k=1}^4(MP_k)^2=\sum_{k=1}^4\overrightarrow{MP_k}^2=\sum_{k=1}^4\overrightarrow{MG +GP_k}^2$$

$$2=\sum_{k=1}^4 \overrightarrow{MG}^2+2\sum_{k=1}^4(\overrightarrow{MG}.\overrightarrow{GP_k})+\sum_{k=1}^4(\overrightarrow{GP_k})^2$$

$$2=4(MG)^2+2 \overrightarrow{MG}.\underbrace{\sum_{k=1}^4 \overrightarrow{GP_k}}_{0}+\underbrace{\sum_{k=1}^4(\overrightarrow{GP_k})^2}_S$$

$$2=4(MG)^2+S \ \text{where} \ S:=\sum_{k=1}^4(\overrightarrow{GP_k})^2 >0,$$

giving a very strong constraint that I would translate into "point $M$ should be very close to centroid $G$".

Jean Marie
  • 81,803