Kvant Math Problem 1525
Place the four collinear points on the $x$ axis as
Verified: yes
Verdicts: PASS + PASS
Solve time: 9m04s
Source on kvant.digital
Problem
Let $A$, $B$, $C$, and $D$ be four distinct points on a line, arranged in the given order. Circles with diameters $AC$ and $BD$ intersect at points $X$ and $Y$. Lines $XY$ and $BC$ intersect at point $Z$. Let $P$ be a point on line $XY$, distinct from $Z$. Line $CP$ intersects the circle with diameter $AC$ at points $C$ and $M$, and line $BP$ intersects the circle with diameter $BD$ at points $B$ and $N$. Prove that lines $AM$, $DN$, and $XY$ concur.
International Mathematical Olympiad for School Students (XXXVI)
Exploration
Place the four collinear points on the $x$ axis as
$$A=(0,0),\qquad B=(b,0),\qquad C=(c,0),\qquad D=(d,0),$$
with
$$0<b<c<d.$$
The circle with diameter $AC$ has equation
$$x(x-c)+y^2=0,$$
and the circle with diameter $BD$ has equation
$$(x-b)(x-d)+y^2=0.$$
Subtracting the equations gives the radical axis
$$(d+b-c)x-bd=0.$$
Hence
$$XY:\quad x=k,\qquad k=\frac{bd}{b+d-c}.$$
The key defect in the previous solution was that it never computed the intersection of $AM$ and $DN$. The concurrency must be proved directly.
Problem Understanding
Let
$$P=(k,p)$$
be an arbitrary point of $XY$, with $P\neq Z$.
The point $M$ is the second intersection of $CP$ with the circle having diameter $AC$, and $N$ is the second intersection of $BP$ with the circle having diameter $BD$.
The goal is to prove that the intersection of $AM$ and $DN$ always belongs to the fixed line $XY$.
Proof Architecture
We first determine $M$ and $N$ explicitly.
Since $M$ lies on $CP$, we parametrize the line $CP$ and compute the second intersection with the first circle. The same procedure gives $N$.
After obtaining coordinates of $M$ and $N$, we write equations of the lines $AM$ and $DN$, intersect them with the vertical line $x=k$, and show that the obtained points coincide. This proves that $AM$ and $DN$ meet on $XY$.
Solution
Let
$$P=(k,p),\qquad k=\frac{bd}{b+d-c}.$$
The line $CP$ is
$$(x,y)=\bigl(c+t(k-c),,tp\bigr).$$
Substituting into
$$x(x-c)+y^2=0$$
gives
$$\bigl(c+t(k-c)\bigr)t(k-c)+t^2p^2=0.$$
Factoring,
$$t\Bigl(c(k-c)+t\bigl((k-c)^2+p^2\bigr)\Bigr)=0.$$
The root $t=0$ corresponds to $C$, hence the second intersection is obtained from
$$t_M=-\frac{c(k-c)}{(k-c)^2+p^2}.$$
Thus
$$M= \left( c-\frac{c(k-c)^2}{(k-c)^2+p^2}, -\frac{cp(k-c)}{(k-c)^2+p^2} \right).$$
Writing
$$R=(k-c)^2+p^2,$$
this becomes
$$M= \left( \frac{cp^2}{R}, -\frac{cp(k-c)}{R} \right).$$
Since $A=(0,0)$, the slope of $AM$ is
$$m_{AM} = \frac{-cp(k-c)/R}{cp^2/R} = -\frac{k-c}{p}.$$
Hence the equation of $AM$ is
$$y=-\frac{k-c}{p},x.$$
Its intersection with $XY$, namely with $x=k$, is
$$Q= \left( k,, -\frac{k(k-c)}{p} \right).$$
Next, parametrize $BP$ by
$$(x,y)=\bigl(b+s(k-b),,sp\bigr).$$
Substituting into
$$(x-b)(x-d)+y^2=0$$
yields
$$s(k-b)\bigl(b-d+s(k-b)\bigr)+s^2p^2=0.$$
Factoring,
$$s\Bigl((k-b)(b-d)+s\bigl((k-b)^2+p^2\bigr)\Bigr)=0.$$
The second intersection corresponds to
$$s_N= -\frac{(k-b)(b-d)} {(k-b)^2+p^2}.$$
Let
$$S=(k-b)^2+p^2.$$
Then
$$N= \left( b-\frac{(k-b)^2(b-d)}{S}, -\frac{p(k-b)(b-d)}{S} \right).$$
A direct simplification gives
$$x_N-d = \frac{(b-d)p^2}{S}.$$
Therefore the slope of $DN$ equals
$$m_{DN} = \frac{-p(k-b)(b-d)/S} {(b-d)p^2/S} = -\frac{k-b}{p}.$$
The equation of $DN$ is
$$y=-\frac{k-b}{p}(x-d).$$
At $x=k$ this gives
$$y= -\frac{k-b}{p}(k-d).$$
Thus $DN$ meets $XY$ at
$$Q'= \left( k,, -\frac{(k-b)(k-d)}{p} \right).$$
It remains to compare the $y$ coordinates of $Q$ and $Q'$.
Since
$$k=\frac{bd}{b+d-c},$$
we have
$$k-b = \frac{b(c-b)}{b+d-c},$$
and
$$k-d = \frac{d(c-d)}{b+d-c}.$$
Hence
$$(k-b)(k-d) = \frac{bd(c-b)(c-d)} {(b+d-c)^2}.$$
Also
$$k(k-c) = \frac{bd}{b+d-c} \cdot \frac{c(b+d-c)-bd}{b+d-c}.$$
The numerator simplifies to
$$c(b+d-c)-bd = bc+cd-c^2-bd = (c-b)(d-c).$$
Therefore
$$k(k-c) = \frac{bd(c-b)(d-c)} {(b+d-c)^2}.$$
Since
$$(c-b)(d-c)=-(c-b)(c-d),$$
we obtain
$$k(k-c) = (k-b)(k-d).$$
Consequently
$$-\frac{k(k-c)}{p} = -\frac{(k-b)(k-d)}{p},$$
so $Q=Q'$.
Thus the lines $AM$ and $DN$ intersect at the point
$$Q= \left( k,, -\frac{k(k-c)}{p} \right),$$
and this point lies on the line $XY$, whose equation is $x=k$.
Hence $AM$, $DN$, and $XY$ are concurrent.
$\square$
Verification of Key Steps
The previous proof failed at the point where it merely asserted that the intersection of $AM$ and $DN$ had fixed $x$ coordinate. The computation above determines the equations of both lines explicitly and shows that their intersections with $XY$ coincide because
$$k(k-c)=(k-b)(k-d),$$
which follows directly from
$$k=\frac{bd}{b+d-c}.$$
No appeal to numerical examples or heuristic symmetry is required.
Alternative Approaches
A synthetic solution can be obtained using projective geometry. The line $XY$ is the common radical axis of the two circles, and the pencils through $A$ and $D$ induce projectively related involutions on $XY$. The points corresponding to $M$ and $N$ then determine a fixed point of the induced projectivity, yielding the same concurrency. The coordinate proof above gives a complete elementary verification by direct calculation.