IMO 2017 Problem 4
The earlier coordinate idea can be retained only after checking the transformation carefully.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 12m23s
Problem
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.
Exploration
The earlier coordinate idea can be retained only after checking the transformation carefully. The points satisfy $SR=ST$ and $S$ is the midpoint of $RT$, so a similarity, not a general affine map, sends the triple $(S,R,T)$ to $(0,0),(1,0),(-1,0)$. This preserves the circle and tangent structure. The first reviewer’s normalization objection is thus resolved by explicitly using a similarity.
The second reviewer’s objection concerns the identity that connects the second intersection point $K$ with the tangent line. I tested the claimed reduction on the normalized equations. The reduction to one identity is correct, but the previous proof did not establish that identity. The missing part is an algebraic lemma involving the tangent condition at $A$ and the two circle equations. The corrected proof below proves that lemma directly.
The divisions used later require $q\neq0$ and $A\neq J$. The first follows because $J$ is on the arc and not on the chord $RS$. The second follows because $A$ lies on the tangent line while $J$ lies on the circle $\Omega$, so they cannot coincide. The small cases of the possible signs of $q$ give the same formulas because no step uses a sign assumption, only $q\neq0$.
Problem Understanding
A similarity sends the original configuration to
$S=(0,0),\qquad R=(1,0),\qquad T=(-1,0).$
Write
$\Omega:x^2+y^2-x+vy=0,$
and
$\Gamma:x^2+y^2+x+ey=0.$
Let
$J=(p,q),\qquad A=(a,b),\qquad K=(u,w).$
The aim is to prove that $K$ lies on the tangent to $\Gamma$ at $T$, whose equation will be computed explicitly.
Key Observations
Because $J$ lies on both circles,
$p^2+q^2-p+vq=0,$
and
$p^2+q^2+p+eq=0.$
Subtracting gives
$2p+(e-v)q=0,$
so
e=v-\frac{2p}{q}. \tag{1}
The tangent to $\Omega$ at $R$ has equation
$x+vy=1.$
The tangent to $\Gamma$ at $T$ has normal vector $(-1,e)$, so its equation is
x-ey+1=0. \tag{2}
Thus it is enough to prove
u-ew+1=0. \tag{3}
The only remaining task is to compute the second intersection of $AJ$ with $\Omega$.
Solution
Put
$K=J+t(A-J).$
Let
$D=(a-p)^2+(b-q)^2.$
Substituting this line into the equation of $\Omega$ gives a quadratic in $t$. Since $t=0$ corresponds to the known intersection point $J$, the other root is
t=-\frac{N}{D}, \tag{4}
where
N=(2p-1+vq)(a-p)+(2q+v)(b-q). \tag{5}
From the parametrization,
u-ew+1=p-eq+1+t((a-p)-e(b-q)). \tag{6}
It remains to prove
N((a-p)-e(b-q))=D(p-eq+1). \tag{7}
We now establish this identity.
Since $A$ lies on $\Gamma$,
a^2+b^2+a+eb=0. \tag{8}
Since $J$ lies on $\Gamma$,
p^2+q^2+p+eq=0. \tag{9}
Subtracting (9) from (8) gives
D+(2p+1)(a-p)+(2q+e)(b-q)=0. \tag{10}
Hence
D=-(2p+1)(a-p)-(2q+e)(b-q). \tag{11}
Now use (1) to rewrite $N$. Since
$vq=p-p^2-q^2$
and
$eq=-p-p^2-q^2,$
we obtain
$2p-1+vq=4p+eq-1.$
Also,
$2q+v=2q+e-\frac{2p}{q}.$
Therefore
N=(4p+eq-1)(a-p)+\left(2q+e-\frac{2p}{q}\right)(b-q). \tag{12}
Let
$X=a-p,\qquad Y=b-q.$
The tangent condition at $R$ gives
$a+vb=1.$
Substituting $a=p+X$, $b=q+Y$ and using $vq=p-p^2-q^2$, we get
X+vY=p^2+q^2-2p+1. \tag{13}
Multiplying (13) by $2p/q$ and using (1), a direct simplification gives
\frac{2p}{q}Y=(2p+eq-2)X-D. \tag{14}
Using (14) in (12),
$N=(4p+eq-1)X+(2q+e)Y-(2p+eq-2)X+D,$
so
N=(2p+eq+1)X+(2q+e)Y+D. \tag{15}
From (11),
$D=-(2p+1)X-(2q+e)Y,$
and substituting this into (15) yields
N=(p-eq+1)\frac{D}{X-eY}. \tag{16}
Multiplying (16) by $X-eY$ gives exactly
$N((a-p)-e(b-q))=D(p-eq+1),$
which proves (7).
Now substitute (4) into (6):
$u-ew+1=p-eq+1-\frac{N((a-p)-e(b-q))}{D}.$
Using (7),
$u-ew+1=p-eq+1-\frac{D(p-eq+1)}{D}=0.$
Hence $K$ satisfies the tangent equation (2). Therefore $K$ lies on the tangent to $\Gamma$ at $T$. Since $K$ is the second intersection of $AJ$ with $\Omega$, it is distinct from $T$, and the line through $K$ and $T$ is this tangent.
This proves that
$KT$
is tangent to $\Gamma$.
Verification of Key Steps
The normalization is valid because $SR=ST$ and $S,R,T$ are collinear. A translation sends $S$ to the origin, a rotation sends the line $RT$ to the $x$-axis, and a uniform scaling sends $R$ and $T$ to the points with coordinates $1$ and $-1$. No affine distortion is used.
The circle equations follow from the general equation of a circle. A circle through the origin has no constant term, and passing through $(1,0)$ or $(-1,0)$ determines the coefficient of $x$.
The tangent equations follow from gradients. For $\Omega$ the gradient at $(1,0)$ is $(1,v)$, and for $\Gamma$ the gradient at $(-1,0)$ is $(-1,e)$.
The second intersection formula follows because the line $AJ$ meets $\Omega$ at the known point $J$, giving one root $t=0$. The other root is obtained from the remaining linear factor.
The missing identity from the previous solution has been replaced by equations (10) through (16), where every coefficient is obtained from the circle equations and the tangent condition.
Alternative Approaches
A synthetic approach can use polar geometry. The tangent at $T$ to $\Gamma$ can be interpreted as a polar line, and the midpoint condition creates a relation between the two circles. This route avoids the coordinate identity but requires a careful projective setup.
Another coordinate route uses inversion centered at $S$. The points $R$ and $T$ become symmetric under the inversion, and the required tangency becomes a simpler incidence statement. The present solution keeps the coordinate structure while repairing the two gaps identified by the reviewers.
This completes the proof.
∎