Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME Find the answer United States

Problem:
Let $ABC$ be a triangle with $AB = 2$ , $AC = 3$ , $BC = 4$ . The isogonal conjugate of a point $P$ , denoted $P^*$ , is the point obtained by intersecting the reflection of lines $PA$ , $PB$ , $PC$ across the angle bisectors of $\angle A$ , $\angle B$ , and $\angle C$ , respectively.

Given a point $Q$ , let $\mathfrak{K}(Q)$ denote the unique cubic plane curve which passes through all points $P$ such that line $PP^*$ contains $Q$ . Consider:

a. the M'Cay cubic $\mathfrak{K}(O)$ , where $O$ is the circumcenter of $\triangle ABC$ ,
b. the Thomson cubic $\mathfrak{K}(G)$ , where $G$ is the centroid of $\triangle ABC$ ,
c. the Napoleon-Feuerbach cubic $\mathfrak{K}(N)$ , where $N$ is the nine-point center of $\triangle ABC$ ,
d. the Darboux cubic $\mathfrak{K}(L)$ , where $L$ is the de Longchamps point (the reflection of the orthocenter across point $O$ ),
e. the Neuberg cubic $\mathfrak{K}(X_{30})$ , where $X_{30}$ is the point at infinity along line $OG$ ,
f. the nine-point circle of $\triangle ABC$ ,
g. the incircle of $\triangle ABC$ , and
h. the circumcircle of $\triangle ABC$ .

Estimate $N$ , the number of points lying on at least two of these eight curves. An estimate of $E$ earns $\left\lfloor 20 \cdot 2^{-|N-E| / 6}\right\rfloor$ points.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
The first main insight is that all the cubics pass through the points $A$ , $B$ , $C$ , $H$ (orthocenter), $O$ , and the incenter and three excenters. Since two cubics intersect in at most nine points, this is all the intersections of a cubic with a cubic.

On the other hand, it is easy to see that among intersections of circles with circles, there are exactly 3 points; the incircle is tangent to the nine-point circle at the Feuerbach point while being contained completely in the circumcircle; on the other hand for this obtuse triangle the nine-point circle and the circumcircle intersect exactly twice.

All computations up until now are exact, so it remains to estimate:
- Intersection of the circumcircle with cubics. Each cubic intersects the circumcircle at an even number of points, and moreover we already know that $A$ , $B$ , $C$ are among these, so the number of additional intersections contributed is either 1 or 3; it is the former only for the Neuberg cubic which has a "loop". Hence the actual answer in this case is $1+3+3+3+3=13$ (but an estimate of $3 \cdot 5=15$ is very reasonable).
- Intersection of the incircle with cubics. Since $\angle A$ is large the incircle is small, but on the other hand we know $I$ lies on each cubic. Hence it's very likely that each cubic intersects the incircle twice (once "coming in" and once "coming out"). This is the case, giving $2 \cdot 5=10$ new points.
- Intersection of the nine-point with cubics. We guess this is close to the 10 points of the incircle, as we know the nine-point circle and the incircle are tangent to each other. In fact, the exact count is 14 points; just two additional branches appear.

In total, $N=9+3+13+10+14=49$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.