Let X,Y be the end points of the diameter of a circumference $\mathit{C}$, and let N be the mid-point of one of the arcs XY of $\mathit{C}$. Let A,B be two points in the segment XY. The lines NA and NB cut $\mathit{C}$ in the points C and D, respectively. The tangents to $\mathit{C}$ in C and D intersect at P. Let M the point of intersection between the segments XY and NP. Prove that M is the mid-point of the segment AB.

I found this exercise on a geometry book (olympiad book without theory, just exercises. I'm learning this type of geometry for the first time) but I do not have any idea on how to tackle it. What theorems could I use to solve this? Any help/hints will be very appreciated.

  • $\begingroup$ The problem was fun to solve. Thanks. What book did you find it in? $\endgroup$
    – Anubhab
    Dec 5 '18 at 9:34
  • $\begingroup$ Unfortunately it's in spanish. It's called "Cuadernos de olimpiada:Geometría" by Radmila Bulajich Manfrino $\endgroup$
    – mobzopi
    Dec 6 '18 at 1:08
  • $\begingroup$ Complex geometry provides a straightforward solution. Despite the fact that you have already accepted an excellent answer from Anubhabh, I decided to post a completely different one. Someone could find it inspiring and useful in similar situations. $\endgroup$
    – Oldboy
    Dec 6 '18 at 8:06


(Please refer to the diagram.) First, we shall prove that $ABDC$ is cyclic. Let $O$ be the center of the original circle. Then, $NO\perp XY$. Therefore, $$NA.NC=NA.AC+NA^2=XA.YA+NO^2+OA^2=(XO-AO)(YO+AO)+NO^2+OA^2=XO^2+NO^2$$ We do the same thing for B and obtain $NB.ND=NA.NC$. Therefore, $ADBC$ is cyclic as claimed.

Let $\angle ANM=\theta_1$, $\angle BNM=\theta_2$, $\angle NAM=\theta_3$, $\angle NBM=\theta_4$. By cyclicity of $ABDC$, $\angle CDN=\theta_3$. As PC is tangent to the circle, $ext. \angle PCN=\theta_3$. Similarly for the angles marked $\theta_4$.

Applying sine rule to $\triangle PCN$ and $\triangle PDN$, we have, $$\frac{PC}{\sin \theta_1}=\frac{PN}{\sin \theta_3}\text { and }\frac{PD}{\sin \theta_2}=\frac{PN}{\sin \theta_4}\text .$$ As $PC=PD$, $$\frac{\sin \theta_1}{\sin \theta_2}=\frac{\sin \theta_3}{\sin \theta_4}\text .$$ Applying sine rule to $\triangle AMN$ and $\triangle BMN$ , we have, $$\frac{AM}{\sin \theta_1}=\frac{MN}{\sin \theta_3}\text { and } \frac{BM}{\sin \theta_2}=\frac{MN}{\sin \theta_4}\text .$$ $$\therefore \frac{AM}{BM}=\frac{\sin\theta_1.\sin\theta_3}{\sin\theta_2.\sin\theta_4}=1$$ $\blacksquare$

Also see: Power of Point, Sine Rule

  • $\begingroup$ Bravo, a really good proof. $\endgroup$
    – Oldboy
    Dec 5 '18 at 22:41
  • 1
    $\begingroup$ Please check my completely different proof. $\endgroup$
    – Oldboy
    Dec 6 '18 at 8:05
  • 1
    $\begingroup$ Thanks. Your complex bash is more straightforward as there is only 1 circle involved. $\endgroup$
    – Anubhab
    Dec 6 '18 at 8:24
  • $\begingroup$ Yes, you have a single circle, but also intersections of chords and tangents and all that pushed me to use complex geometry. $\endgroup$
    – Oldboy
    Dec 6 '18 at 8:59
  • 1
    $\begingroup$ Fixed it. Thanks. $\endgroup$
    – Anubhab
    Dec 6 '18 at 9:42

This problem can be solved in a completely straightforward don't-make-me-think way by using complex geometry (which is a frequent subject in IMO and other competitions). The most important formulas can be found HERE.

WLOG, we can assume that our circle is a unit circle in a complex plane:

enter image description here

Each point in the drawing is represented with a complex number. A capital letter (say $R$) is a point, the corresponding small letter ($r$) is its complex coordinate (usual convention in complex geometry).

We'll pick points $C$ and $D$ freely on the unit circle. These two points satisfy the following relation:

$$\bar{c}=\frac1c, \ \bar{d}=\frac1d\tag{1}$$

The point $A$ represents the intersection of chords $NC$ and $XY$. The point of intersection is given with the following (well-known) formula (also found in the "cheat sheet" mentioned above):


Notice that $n=1$, $x=i$, $y=-i$, $x+y=0$, $xy=1$. This gives:


In a similar fashion:


Denote the midpoint of AB with M:


Point P is defined by the intersection of tangents $CP$ and $CD$. Again, by a well-known formula:


Let us prove that points $N,M,P$ are collinear and we are done! In complex geometry, this is true iff:


Obviously $n=\bar{n}=1$. So we have to prove that:


From (2):



From (3):



By comparing (5) and (6) we see that (4) is true and therefore points $N,M,P$ must be collinear.


  • $\begingroup$ Are you a IMO competitor? $\endgroup$
    – Anubhab
    Dec 6 '18 at 8:29
  • $\begingroup$ No, I'm way too old for it :) $\endgroup$
    – Oldboy
    Dec 6 '18 at 8:54
  • $\begingroup$ I assume you are a trainer, then. $\endgroup$
    – Anubhab
    Dec 6 '18 at 8:56
  • 1
    $\begingroup$ @AnubhabGhosal No, I just love math and learned a bit of it over decades. I trained my son until he reached the level to start training me. :) $\endgroup$
    – Oldboy
    Dec 6 '18 at 9:01

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