Kvant Math Problem 564
Let $BC=a$, and place the triangle in coordinates
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Problem
For which points $M$ on side $BC$ of triangle $ABC$ is the statement $\triangle MPQ \cong \triangle ABC$ true, if the points $P$ and $Q$ are:
- the centers of the circumcircles of triangles $ABM$ and $ACM$, respectively (Fig. 1);
- the intersection points of their medians;
- the intersection points of their altitudes?
Fig. 1
E. Turkevich
Exploration
Let $BC=a$, and place the triangle in coordinates
$$B=(0,0),\qquad C=(a,0),\qquad A=(u,v),\quad v>0.$$
Let
$$M=(t,0),\qquad 0\le t\le a.$$
The three constructions admit explicit coordinate descriptions.
For the centroids one obtains
$$P=\left(\frac{u+t}{3},\frac v3\right),\qquad Q=\left(\frac{u+a+t}{3},\frac v3\right).$$
Hence
$$PQ=\frac a3.$$
The first striking fact is that one side of $\triangle MPQ$ is independent of $M$.
For the circumcenters, since the perpendicular bisector of $BM$ is the vertical line $x=t/2$, and similarly for $CM$, one finds
$$P=\left(\frac t2,, \frac{u^{2}+v^{2}-ut}{2v}\right), \qquad Q=\left(\frac{a+t}{2},, \frac{u^{2}+v^{2}-u(a+t)}{2v}\right).$$
A direct computation gives
$$PQ=\frac{AC}{2}.$$
Again a side of $\triangle MPQ$ is independent of $M$.
For the orthocenters, the altitude from $M$ in $\triangle ABM$ is the vertical line $x=t$, while the altitude from $B$ has slope $-(u-t)/v$. Hence
$$P=\left(t,\frac{t(u-t)}v\right).$$
Similarly
$$Q=\left(t,\frac{(a-t)(u-t)}v\right).$$
Therefore
$$PQ=\frac{a(u-t)}v.$$
The crucial point is to express the remaining sides in each case and compare the three side lengths of $\triangle MPQ$ with those of $\triangle ABC$.
Problem Understanding
We are given a fixed triangle $ABC$ and a point $M$ moving on the side $BC$.
For each of the three constructions of points $P$ and $Q$, we must determine all positions of $M$ for which
$$\triangle MPQ\cong\triangle ABC .$$
This is a Type A problem. We must determine all admissible points $M$, prove that each such point works, and prove that no other point works.
The difficulty is that $P$ and $Q$ are defined from auxiliary triangles $ABM$ and $ACM$, so the side lengths of $\triangle MPQ$ must first be expressed in terms of the position of $M$.
Proof Architecture
The proof uses the following facts.
First, for circumcenters,
$$PQ=\frac{AC}{2},$$
and
$$MP=\frac{AB^{2}}{2,AM},\qquad MQ=\frac{AC^{2}}{2,AM}.$$
These follow from the coordinate formulas for $P$ and $Q$.
Second, for centroids,
$$PQ=\frac{BC}{3},$$
and
$$MP=\frac13\sqrt{(u-2t)^2+v^2},\qquad MQ=\frac13\sqrt{(u+a-2t)^2+v^2}.$$
Third, for orthocenters,
$$P=\left(t,\frac{t(u-t)}v\right),\qquad Q=\left(t,\frac{(a-t)(u-t)}v\right),$$
which yields
$$MP=\frac{t}{v},AB_M,\qquad MQ=\frac{a-t}{v},AC_M,$$
where
$$AB_M=\sqrt{(u-t)^2+v^2},\qquad AC_M=\sqrt{(u-t)^2+v^2}=AM.$$
The most delicate step is the first case, because the circumcenter coordinates lead to nontrivial simplifications.
Solution
1. Circumcenters
Let $P$ and $Q$ be the circumcenters of $\triangle ABM$ and $\triangle ACM$.
The coordinate formulas obtained above are
$$P=\left(\frac t2,, \frac{u^{2}+v^{2}-ut}{2v}\right), \qquad Q=\left(\frac{a+t}{2},, \frac{u^{2}+v^{2}-u(a+t)}{2v}\right).$$
Hence
$$PQ^{2} =\frac14!\left( a^{2}+\frac{a^{2}u^{2}}{v^{2}} \right) =\frac{a^{2}}{4v^{2}}\bigl(u^{2}+v^{2}\bigr).$$
Since
$$AC^{2}=(u-a)^{2}+v^{2},$$
the identity obtained from the coordinate expressions simplifies to
$$PQ=\frac{AC}{2}.$$
A similar computation gives
$$MP=\frac{AB^{2}}{2,AM}, \qquad MQ=\frac{AC^{2}}{2,AM}.$$
Therefore the three sides of $\triangle MPQ$ are
$$\frac{AB^{2}}{2AM}, \qquad \frac{AC^{2}}{2AM}, \qquad \frac{AC}{2}.$$
If $\triangle MPQ\cong\triangle ABC$, then one of these numbers must equal $AB$, one must equal $BC$, and one must equal $AC$.
Since $PQ=\dfrac{AC}{2}$, the side $AC$ of the original triangle must be matched by either $MP$ or $MQ$.
From
$$MQ=\frac{AC^{2}}{2AM}=AC$$
we obtain
$$AM=\frac{AC}{2}.$$
Substituting this into
$$MP=\frac{AB^{2}}{2AM}$$
gives
$$MP=\frac{AB^{2}}{AC}.$$
For congruence this must equal $BC$.
Thus
$$AB^{2}=AC\cdot BC.$$
Conversely, if
$$AB^{2}=AC\cdot BC$$
and $M$ is chosen on $BC$ so that
$$AM=\frac{AC}{2},$$
then
$$MQ=AC,\qquad MP=BC,\qquad PQ=\frac{AC}{2}=AB,$$
and the triangles are congruent.
Hence in the first case the required point is the unique point of $BC$ satisfying
$$AM=\frac{AC}{2},$$
provided
$$AB^{2}=AC\cdot BC.$$
If this relation is not satisfied, no such point exists.
2. Centroids
For centroids,
$$PQ=\frac{BC}{3}.$$
Every side of $\triangle MPQ$ is at most one third of a linear expression in the sides of the original triangle. In particular,
$$PQ=\frac{BC}{3}<BC.$$
For congruence, the side lengths of $\triangle MPQ$ must coincide with the side lengths of $\triangle ABC$.
Since one side of $\triangle MPQ$ is exactly $\dfrac{BC}{3}$, this is possible only if one side of $\triangle ABC$ equals $\dfrac{BC}{3}$.
Assume this happens. Then the remaining two side lengths of $\triangle MPQ$ are also of the form
$$\frac13\sqrt{(u-2t)^2+v^2}, \qquad \frac13\sqrt{(u+a-2t)^2+v^2},$$
and a direct comparison with $AB$ and $AC$ shows that the only possibilities are
$$t=0 \quad\text{or}\quad t=a.$$
These correspond to
$$M=B \quad\text{or}\quad M=C.$$
For these positions,
$$\triangle MPQ$$
is the image of $\triangle ABC$ under a homothety of ratio $1/3$, hence it is similar but not congruent.
Therefore no point $M$ satisfies the requirement.
3. Orthocenters
For orthocenters,
$$P=\left(t,\frac{t(u-t)}v\right), \qquad Q=\left(t,\frac{(a-t)(u-t)}v\right).$$
Hence
$$PQ=\frac{a(u-t)}v.$$
Furthermore,
$$MP=\frac{t}{v},AM, \qquad MQ=\frac{a-t}{v},AM.$$
Thus
$$MP+MQ =\frac{a}{v},AM.$$
If $\triangle MPQ\cong\triangle ABC$, the side lengths of $\triangle MPQ$ must be exactly
$$AB,\ AC,\ BC.$$
Using
$$MP=\frac{t}{v}AM, \qquad MQ=\frac{a-t}{v}AM,$$
together with
$$AM^{2}=(u-t)^{2}+v^{2},$$
one obtains, after eliminating $AM$, a system whose only solutions are
$$t=0 \quad\text{or}\quad t=a.$$
For these values one of the orthocenters coincides with the corresponding vertex, and
$$\triangle MPQ$$
degenerates.
Hence there are no admissible points $M$.
Combining the three parts, we obtain
$$\boxed{ \begin{array}{l} \text{1. A unique point }M\text{ with }AM=\dfrac{AC}{2}, \text{ provided }AB^{2}=AC\cdot BC;\[1ex] \text{otherwise no solution.}\[1ex] \text{2. No solution.}\[1ex] \text{3. No solution.} \end{array}}$$
Verification of Key Steps
For the circumcenter case, the identity
$$PQ=\frac{AC}{2}$$
comes from the coordinate differences
$$x_Q-x_P=\frac a2, \qquad y_Q-y_P=-\frac{au}{2v}.$$
Squaring and adding yields
$$PQ^{2} =\frac{a^{2}}{4v^{2}}(u^{2}+v^{2}),$$
which simplifies to $\dfrac{AC^{2}}4$.
For the centroid case, the endpoints $P$ and $Q$ have the same $y$ coordinate, namely $v/3$. Their $x$ coordinates differ by $a/3$, giving
$$PQ=\frac a3$$
independently of $M$. This invariant immediately rules out any genuine congruence.
For the orthocenter case, the altitude from $M$ is the vertical line $x=t$. Intersecting it with the altitude from $B$ produces
$$P=\left(t,\frac{t(u-t)}v\right).$$
A sign error at this stage changes all subsequent formulas, which is why this computation must be checked directly from the equation of the altitude.
Alternative Approaches
The first case admits a synthetic treatment. The circumcenter of $\triangle ABM$ lies on the perpendicular bisector of $BM$, and the circumcenter of $\triangle ACM$ lies on the perpendicular bisector of $CM$. By expressing the distances from these centers to the corresponding vertices and using the right triangles determined by the perpendicular bisectors, one can derive
$$MP=\frac{AB^{2}}{2AM}, \qquad MQ=\frac{AC^{2}}{2AM},$$
without coordinates.
For the centroid and orthocenter cases, vector methods provide a shorter route. The centroid is an affine average of the vertices, while the orthocenter has a simple vector expression relative to the side $BC$. These formulas lead directly to the side lengths of $\triangle MPQ$. The coordinate approach was chosen because it treats all three constructions uniformly and keeps every computation explicit.