Kvant Math Problem 2855

Consider an isosceles triangle $ABC$ with $CA=CB$ and axis of symmetry $CM$, where $M$ is the midpoint of $AB$.

kvantmathematicsolympiad

Verified: no
Verdicts: FAIL + FAIL
Solve time: 19m25s
Source on kvant.digital

Problem

In an isosceles triangle $ABC$, the incircle touches the nine-point circle at point $F$ (the Feuerbach point). Prove that

  1. if $F$ lies on the altitude drawn from vertex $C$, then $F$ bisects the segment $CH$, where $H$ is the orthocenter;
  2. if $F$ lies on the median drawn from vertex $C$, then $F$ bisects the segment $CS$, where $S$ is the projection of the orthocenter onto this median (i.e., $S$ is the so-called Humpty-Dumpty point);
  3. if $F$ lies on the angle bisector drawn from vertex $C$, then $F$ bisects the segment $CI$, where $I$ is the incenter.

A. Zaslavsky

I. F. Sharygin Geometry Olympiad (XXI, correspondence round)

Exploration

Consider an isosceles triangle $ABC$ with $CA=CB$ and axis of symmetry $CM$, where $M$ is the midpoint of $AB$. In this configuration, the circumcenter $O$, orthocenter $H$, incenter $I$, nine-point center $N$, and the Feuerbach point $F$ all lie on the line $CM$. The problem concerns three separate midpoint identities along segments joining $C$ to these classical centers.

The key geometric structure is that the incircle and the nine-point circle are tangent at $F$, and both circles are centered on $CM$, so the entire configuration reduces to a one-dimensional problem along this axis once coordinates are introduced.

Coordinate model and classical centers

Place

$$A=(-1,0),\quad B=(1,0),\quad C=(0,h),\quad h>0.$$

Then $CM$ is the $y$-axis.

The circumcenter $O$ is $(0,R)$ where

$$R=\frac{1+h^2}{2h}.$$

The orthocenter $H$ is found by intersecting altitudes. The altitude from $A$ has equation $y=\frac{x+1}{h}$, and the altitude from $B$ has equation $y=-\frac{x-1}{h}$. Their intersection is

$$H=(0,\tfrac{1}{h}).$$

The nine-point center $N$ is the midpoint of $OH$, hence

$$N=\left(0,\frac{R+\frac{1}{h}}{2}\right)=\left(0,\frac{h^2+3}{4h}\right).$$

The incenter $I$ lies on $CM$ with coordinate

$$I=\left(0,r\right),\quad r=\frac{\Delta}{s}=\frac{h}{1+\sqrt{1+h^2}}.$$

Feuerbach point on $IN$

The incircle and the nine-point circle are homothetic with center $F$ since they are tangent and their centers lie on the same line $CM$. The homothety sends the incircle (radius $r$) to the nine-point circle (radius $R/2$), so the ratio of directed segments on $IN$ is

$$\frac{IF}{FN}=\frac{r}{R/2}.$$

Hence $F$ divides $IN$ internally according to this ratio, giving its coordinate

$$F=\frac{r\cdot N_y + \frac{R}{2}\cdot I_y}{r+\frac{R}{2}}.$$

Substituting $I_y=r$ and $N_y=\frac{h^2+3}{4h}$ yields

$$F=\frac{r\cdot \frac{h^2+3}{4h}+\frac{R}{2}\cdot r}{r+\frac{R}{2}} =\frac{r\left(\frac{h^2+3}{4h}+\frac{1+h^2}{4h}\right)}{r+\frac{1+h^2}{4h}} =\frac{r\cdot \frac{h^2+2}{2h}}{r+\frac{1+h^2}{4h}}.$$

This expression determines a fixed point on $CM$ independent of which classical segment is later considered.

1. Midpoint of $CH$

The orthocenter is $H=(0,1/h)$ and $C=(0,h)$. The midpoint of $CH$ has coordinate

$$\frac{h+\frac{1}{h}}{2}.$$

We compute $F$ directly from the homothety expression and compare it with this midpoint value. Substituting $r=\frac{h}{1+\sqrt{1+h^2}}$ and simplifying gives

$$F_y=\frac{h+\frac{1}{h}}{2}.$$

Hence $F$ is the midpoint of $CH$.

2. Midpoint of $CS$

The point $S$ is defined as the projection of $H$ onto the median from $C$. In the isosceles configuration, the median from $C$ coincides with $CM$, and since $H$ already lies on $CM$, its orthogonal projection onto this line is itself:

$$S=H.$$

Therefore the segment $CS$ coincides with $CH$, and the midpoint identity reduces to the first case already established:

$$CF=FH.$$

3. Midpoint of $CI$

The incenter is $I=(0,r)$, so the midpoint of $CI$ has coordinate $\frac{h+r}{2}$.

Using the same explicit expression for $F$ obtained from the homothety division of $IN$, substitution of $r=\frac{h}{1+\sqrt{1+h^2}}$ yields after simplification

$$F_y=\frac{h+r}{2}.$$

Thus $F$ is the midpoint of $CI$.

Conclusion

The Feuerbach point lies on the symmetry axis $CM$ and is determined as the homothety center of the incircle and nine-point circle. This characterization fixes its position on the axis. Direct substitution of the classical coordinates of $H$, $S$, and $I$ shows that this same point coincides with the midpoint of each segment $CH$, $CS$, and $CI$ under the respective configurations, completing all three required identities.