There are $n$ different points $A_1, \ldots , A_n$ in the plain and each point $A_i$ it is assigned a real number $\lambda_i$ distinct from zero in such way that $(\overline{A_i A_j})^2 = \lambda_i + \lambda_j$ for all the $i$,$j$ with $i\neq{}j$} Show that: (1) $n \leq 4$ (2) If $n=4$, then $\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}+ \frac{1}{\lambda_4} = 0$
Problem
Source: Spanish Communities
Tags: geometry, circumcircle, trigonometry, geometry unsolved
Xixas
23.04.2006 12:01
Here is my solution for this problem: At first we will show that $n\le 4$, so for sake of contradiction let $n>4$. We will be using the criterion for perpendicular segments, namely the one that $AB\bot CD\Leftrightarrow AC^2-BC^2=AD^2-BD^2$. It is trivial using Pythagoras' theorem. Let us suppose that some three of the points ($A_1, A_2, A_3$) lie on a single line. Then if all points lie on this line without loss of generality we may assume the line is horizontal and they are ordered from left to right $A_1, A_2, \ldots, A_n$. We see that \[ A_1A_{n-1}^2-A_nA_{n-1}^2=\lambda_1+\lambda_{n-1}-\lambda_n-\lambda_{n-1}=\lambda_1+\lambda_{2}-\lambda_n-\lambda_{2}= A_1A_{2}^2-A_nA_{2}^2<A_1A_{n-1}^2-A_nA_{n-1}^2 \] and we get a contradiction since $A_2 \neq A_{n-1}$. So let $A_{4}$ be a point that does not lie on that line. We get $A_1A_4^2-A_1A_2^2=\lambda_1+\lambda_4-\lambda_1-\lambda_2=\lambda_3+\lambda_4-\lambda_3-\lambda_2=A_3A_4^2-A_3A_{2}^2$ and hence $A_1A_3\bot A_2A_4$. Analogously we obtain $A_1A_2\bot A_3A_4$ which gives a contradiction. Hence no three of our points are collinear. But noting that for fixed $A_1, A_2, A_3$ and distinct from them $A_i$ in the same way as above we get $A_1A_2\bot A_3A_i$ and since no three of the points are collinear we get that $n\le 4$. Next suppose that $n=4$. Say our four points $A_1, A_2, A_3, A_4$ form a convex quadrilateral (with perpendicular diagonals as seen from the proof above). If the feet of a perpendicular from $A_1$ to $A_3A_4$ does not belong to $\overline{A_3A_4}$ then we easily get that all such feets of perpendiculars do not lie on respective sides of a quadrilateral and thus all the angles of a quadrilateral are $>90^o$ which is impossible. But in the other case we get that we have a square $A_1A_2A_3A_4$ for which $\lambda_1=\lambda_3=-\lambda_2=-\lambda_4$ and the result follows. In the case when $A_1A_2A_3A_4$ is not convex using the criterion we get that $A_i$ is an orthocenter of the triangle formed by the other three points. Alas I do not know how to finish the proof in this case.
Jorge Miranda
13.08.2009 21:00
Lemma: If $ A,B,C,D$ are distinct points, then $ (AB^2+CD^2=BC^2+DA^2) \Leftrightarrow (AC\perp BD)$
ProofWe have that:
$ AB^2+CD^2=BC^2+DA^2\Leftrightarrow (A-B)^2+(C-D)^2=(B-C)^2+(D-A)^2 \Leftrightarrow$
$ \Leftrightarrow AB+CD=BC+DA \Leftrightarrow$
$ \Leftrightarrow (A-C)(B-D)=0 \Leftrightarrow AC \perp BD$ (since $ A\neq C$, $ B\neq D$)
Let $ 1\leq i,j,k,l\leq n$ be distinct integers, and assume that $ n\geq 4$.
$ A_iA_j^2+A_kA_l^2=(\lambda_i+\lambda_j)+(\lambda_k+\lambda_l)=(\lambda_i+\lambda_k)+(\lambda_j+\lambda_l)=A_iA_k^2+A_jA_l^2$
Using the lemma, we have that $ A_iA_l \perp A_kA_j$. Analogously we get $ A_jA_l\perp A_iA_k$, $ A_kA_l\perp A_iA_j$, so $ A_l$ is the orthocenter of $ \triangle A_iA_jA_k$. In particular, since the orthocenter of a triangle is unique, we get that $ n\leq 4$.
Now let the points be $ A,B,C,H$ with $ H$ in the interior of $ \triangle ABC$, $ BC=a,CA=b,AB=c$, and $ R$ is the circumradius of $ \triangle ABC$.
$ \lambda_A+\lambda_B=AB^2=c^2$
$ \lambda_A+\lambda_C=AC^2=b^2$
Subtracting, $ \lambda_B-\lambda_C=c^2-b^2$. But $ \lambda_B+\lambda_C=BC^2=a^2$, so $ \lambda_B=\frac{a^2+c^2-b^2}{2}=ac\cos(\beta)$, by the cosine law, and analogous equalities for $ A,C,H$.
By the sine law, $ \frac{AH}{AB}=\frac{\sin(\frac{\pi}{2}-\alpha)}{\sin(\gamma)} \Leftrightarrow AH=2R\cos(\alpha)$, and $ BH=2R\cos(\beta)$.
We have to prove that:
$ \frac{1}{\lambda_A}+\frac{1}{\lambda_B}+\frac{1}{\lambda_C}+\frac{1}{\lambda_H}= 0\Leftrightarrow$
$ \Leftrightarrow \sum_{cyc}\frac{1}{ab\cos(\gamma)}=-\frac{1}{AH\times BH \cos(\pi-\gamma)}=\frac{1}{AH\times BH \cos (\gamma)}\Leftrightarrow$
$ \Leftrightarrow \sum_{cyc}\frac{1}{4R^2\sin(\alpha)\sin(\beta)\cos(\gamma)}=\frac{1}{4R^2 \cos (\alpha)\cos(\beta)\cos (\gamma)}\Leftrightarrow$
$ \Leftrightarrow \tan(\alpha)+\tan(\beta)+\tan(\gamma)=\tan(\alpha)\tan(\beta)\tan(\gamma)$
This last equality is true because $ \alpha+\beta+\gamma=\pi$, as desired.
parmenides51
11.12.2022 01:37
There are $n$ different points $A_1,... , A_n$ in the plane and at each point $A_i$ is assigned a real number $\lambda_i$, distinct from zero, in such way that $(\overline{A_i A_j})^2 = \lambda_i + \lambda_j$ for all $i,j$ with $i\ne j$ . Show that: (1) $n \le 4$ (2) If $n=4$, then $\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}+ \frac{1}{\lambda_4} = 0$