2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
3
$\begingroup$

Diagram

(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

$\endgroup$
6
  • $\begingroup$ 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
1
$\begingroup$

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):

$$a=\frac{nc(x+y)-xy(n+c)}{nc-xy}$$

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

$$a=\frac{1+c}{1-c}$$

In a similar fashion:

$$b=\frac{1+d}{1-d}$$

Denote the midpoint of AB with M:

$$m=\frac{a+b}2=\frac{1-cd}{(1-c)(1-d)}\tag{2}$$

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

$$p=\frac{2cd}{c+d}\tag{3}$$

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

$$\frac{n-m}{\bar{n}-\bar{m}}=\frac{n-p}{\bar{n}-\bar{p}}$$

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

$$\frac{1-m}{1-\bar{m}}=\frac{1-p}{1-\bar{p}}\tag{4}$$

From (2):

$$\bar{m}=\frac{1-\bar{c}\bar{d}}{(1-\bar{c})(1-\bar{d})}=\frac{1-\frac1c\frac1d}{(1-\frac1c)(1-\frac1d)}=\frac{cd-1}{(c-1))(d-1)}$$

$$\frac{1-m}{1-\bar{m}}=\frac{1-\frac{1-cd}{(1-c)(1-d)}}{1-\frac{cd-1}{(c-1))(d-1)}}=\frac{c+d-2cd}{c+d-2}\tag{5}$$

From (3):

$$\bar{p}=\frac{2\bar{c}\bar{d}}{\bar{c}+\bar{d}}=\frac{2\frac1c\frac1d}{\frac1c+\frac1d}=\frac{2}{c+d}$$

$$\frac{1-p}{1-\bar{p}}=\frac{1-\frac{2cd}{c+d}}{1-\frac{2}{c+d}}=\frac{c+d-2cd}{c+d-2}\tag{6}$$

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

Done.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.