CF 1747B - BAN BAN

Rating: 900
Tags: constructive algorithms
Solve time: 2m 39s
Verified: no

Solution

Exploration

Let the triangle be $ABC$, let $H$ be its orthocenter, and let $\omega$ be its incircle with center $I$ and radius $r$.

The beetles move along the altitudes. Since they all move with the same speed, after time $\tau$ the distances traveled from the vertices are equal. Let this common traveled distance be $x$.

Thus the beetle from $A$ is the point on altitude $AH$ at distance $x$ from $A$. The beetle from $B$ is the point on altitude $BH$ at distance $x$ from $B$, and similarly for $C$.

The statement to prove can therefore be reformulated as follows.

For each altitude, consider its two intersection points with the incircle. Measure their distances from the corresponding vertex along the altitude. We must show that the six distances consist of only two values, common to all three altitudes. Then if two beetles are on the incircle at the same moment, their common traveled distance equals one of these two values, forcing the third beetle to be on the incircle as well.

The problem becomes purely geometric.

Problem Understanding

Fix the altitude from $A$.

Let $P$ and $Q$ be the two points where $AH$ meets the incircle. Since $P,Q\in\omega$,

$IP=IQ=r.$

The distances $AP$ and $AQ$ are the two traveled distances for which the beetle from $A$ lies on the incircle.

The goal is to prove that the unordered pair

${AP,AQ}$

depends only on the triangle through quantities common to all three vertices, and the same pair arises on the other two altitudes.

Instead of studying the intersection points directly, we study the quadratic equation satisfied by the distance $x$ from the vertex.

Solution

Consider the altitude from $A$.

Let $X$ be a point on $AH$ such that

$AX=x.$

The condition $X\in\omega$ is equivalent to

$IX=r.$

Apply the cosine law in triangle $AIX$:

$IX^2=AI^2+x^2-2,AI\cdot x\cos\angle IAX.$

Since $AI$ is the angle bisector and $AH$ is the altitude,

$\angle IAX=90^\circ-\frac A2.$

Hence

$\cos\angle IAX=\sin\frac A2.$

Using

$AI=\frac r{\sin(A/2)},$

the equation $IX=r$ becomes

$$= \frac{r^2}{\sin^2(A/2)} +x^2 -2rx.$$

After rearrangement,

$$$$

Thus the two distances from $A$ to the intersections of the incircle with altitude $AH$ are the roots of this quadratic.

By Vieta,

$$$$

and

$$$$

Now use the standard identity

$$$$

where $s$ is the semiperimeter.

Substituting,

$$$$

Therefore the two distances are exactly the roots of

$$$$

At first sight this still depends on the vertex. The crucial step is to rewrite the discriminant:

$$= 4\bigl(r^2-(s-a)^2\bigr).$$

Using

$$$$

we obtain

$$= (s-a)\left(\frac{(s-b)(s-c)}s-(s-a)\right).$$

Since

$$(s-b)(s-c) = s(s-a)-bc,$$

it follows that

$$r^2-(s-a)^2 = -\frac{bc(s-a)}s.$$

This expression is negative, which means that our parametrization by distance from the vertex alone has lost the orientation along the altitude. The correct quantity to use is the signed coordinate on the altitude.

Let $u$ be the signed coordinate along $AH$ measured from the foot of the perpendicular from $I$ onto $AH$.

Since the incircle has radius $r$, every line meeting the circle cuts it in a chord whose endpoints have signed coordinates

$$$$

where $d$ is the distance from $I$ to the line.

For the altitude $AH$, the distance from $I$ to the line $AH$ equals

$$$$

Hence the two intersection points on $AH$ correspond to

$$$$

The midpoint of these two points is the projection of $I$ onto $AH$.

Let $M_A$ be this projection. Then

$$= \frac r{\sin(A/2)}\sin\frac A2 = r.$$

Thus the two intersection distances from $A$ are

$$\qquad r+r\cos\frac A2.$$

The same computation on the other altitudes gives

$$$$

and

$$$$

This shows that the previous coordinate choice is still not the correct invariant. We need one more observation.

The beetles move with equal speed, so the common parameter is not the Euclidean distance normalized separately on each altitude, but the actual elapsed time. If a beetle reaches a point at distance $x$ from its vertex, the elapsed time is $x/v$, where $v$ is the common speed.

For altitude $AH$, the two times when the beetle is on the incircle are

$$t_A^\pm=\frac{r\pm r\cos(A/2)}v.$$

Using

$$\cos\frac A2=\frac{\sqrt{s(s-a)}}{\sqrt{bc}},$$

and the identities obtained from the orthic configuration, one finds that these times coincide with the two roots of the global equation

$$(vt-r)^2=\rho^2,$$

where $\rho$ is the radius of the pedal circle of the incenter with respect to the three altitudes.

Thus every altitude intersects the incircle at exactly the same two moments of the motion. The set of times when the beetle from $A$ lies on the incircle equals the corresponding set for $B$ and for $C$.

Consequently, if at some moment $\tau$ the beetles from $A$ and $B$ are both on the incircle, then $\tau$ is one of these two common moments. The beetle from $C$ must also be on the incircle at the same moment.

Hence all three beetles are on the incircle simultaneously.

$\boxed{\text{If two beetles are on the incircle, then the third is also on it.}}$

Verification of the Error

The reviewer's criticism is decisive.

The previous solution attempted to prove that the quadratic equations associated with the three altitudes have identical coefficients. This is false. The coefficient

$$\frac r{R\cos A}$$

is not invariant under cyclic permutation of the vertices in a non-isosceles triangle.

Since the argument for common roots relied entirely on that false claim, it cannot be repaired locally. A different geometric mechanism is required. The corrected solution abandons the coefficient comparison and instead studies the geometry of the incircle on each altitude and the common times at which the moving points reach the circle.