Kvant Math Problem 213
Let the circle have center $I$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m31s
Source on kvant.digital
Problem
Given an angle with vertex $O$ and a circle tangent to its sides at points $A$ and $B$. From point $A$, a ray is drawn parallel to $OB$, intersecting the circle at point $C$. Segment $OC$ intersects the circle at point $E$, and the lines $AE$ and $OB$ intersect at point $K$. Prove that $OK = KB$.
E. V. Sallinen
All-Union Mathematical Olympiad for School Students (VII, Grades 9 and 10)
Exploration
Let the circle have center $I$. Since the circle is tangent to the sides of the angle at $A$ and $B$, the radii $IA$ and $IB$ are perpendicular to $OA$ and $OB$ respectively.
A convenient coordinate model suggests itself. Put the circle as the unit circle
$$x^2+y^2=1$$
with
$$A=(0,1),\qquad B=(1,0).$$
Then the tangents at $A$ and $B$ are the lines $y=1$ and $x=1$, whose intersection is
$$O=(1,1).$$
The ray from $A$ parallel to $OB$ has direction
$$\overrightarrow{OB}=(0,-1),$$
so it is the vertical line $x=0$. Its second intersection with the circle is
$$C=(0,-1).$$
The line $OC$ joins $(1,1)$ and $(0,-1)$, hence
$$y=2x-1.$$
Its intersections with the circle satisfy
$$x^2+(2x-1)^2=1,$$
giving
$$5x^2-4x=0.$$
One root is $x=0$, corresponding to $C$; the other is
$$x=\frac45,\qquad y=\frac35.$$
Thus
$$E=\left(\frac45,\frac35\right).$$
Now compute the line $AE$. Its slope is
$$\frac{\frac35-1}{\frac45}=-\frac12,$$
so
$$AE:\quad y=1-\frac{x}{2}.$$
Intersecting with $OB$, which is $x=1$, yields
$$K=\left(1,\frac12\right).$$
Since
$$O=(1,1),\qquad B=(1,0),$$
the point $K$ is the midpoint of $OB$. Hence
$$OK=KB.$$
The computation is very short and suggests that the essential fact is that, in suitable coordinates, $AE$ always passes through the midpoint of $OB$. The step most likely to conceal an error is the determination of $E$ as the second intersection of $OC$ with the circle; this must be checked carefully.
Problem Understanding
We are given an angle with vertex $O$ and an inscribed circle tangent to its sides at $A$ and $B$. Through $A$ a ray parallel to $OB$ meets the circle again at $C$. The line $OC$ meets the circle again at $E$. The lines $AE$ and $OB$ meet at $K$. We must prove that $K$ is the midpoint of $OB$.
This is a Type B problem, a pure proof.
The core difficulty is to relate the several auxiliary intersections $C$, $E$, and $K$ created from the tangency configuration. A coordinate system adapted to the circle and the tangents turns all constructions into explicit linear equations.
Proof Architecture
First, choose coordinates so that the circle is the unit circle, the tangency points are $A=(0,1)$ and $B=(1,0)$, and the vertex of the angle is $O=(1,1)$; this follows because the sides of the angle are the tangents at $A$ and $B$.
Second, determine $C$ as the second intersection of the line through $A$ parallel to $OB$ with the circle; this gives $C=(0,-1)$.
Third, determine $E$ as the second intersection of the line $OC$ with the circle; solving the resulting system yields $E=(4/5,3/5)$.
Fourth, compute the equation of $AE$ and intersect it with $OB$; this gives $K=(1,1/2)$.
Finally, since $O=(1,1)$ and $B=(1,0)$, the point $(1,1/2)$ is the midpoint of $OB$, hence $OK=KB$.
The lemma most likely to fail under scrutiny is the computation of the second intersection point $E$, because choosing the wrong root would invalidate the remainder of the argument.
Solution
Let the given circle be represented in Cartesian coordinates by
$$x^2+y^2=1.$$
Choose coordinates so that
$$A=(0,1),\qquad B=(1,0).$$
The tangent to the circle at $A$ is the line $y=1$, and the tangent at $B$ is the line $x=1$. Since the sides of the angle are precisely these tangents, their intersection is
$$O=(1,1).$$
The line $OB$ is the vertical line $x=1$. The ray through $A$ parallel to $OB$ is therefore the line $x=0$. Its second intersection with the unit circle is
$$C=(0,-1).$$
The line through $O=(1,1)$ and $C=(0,-1)$ has equation
$$y=2x-1.$$
To find its intersections with the circle, substitute into
$$x^2+y^2=1:$$
$$x^2+(2x-1)^2=1.$$
Hence
$$5x^2-4x=0,$$
so
$$x=0$$
or
$$x=\frac45.$$
The first solution corresponds to the point $C$. Thus the second intersection point is
$$E=\left(\frac45,\frac35\right).$$
Now compute the equation of the line $AE$. Its slope equals
$$\frac{\frac35-1}{\frac45} = -\frac12.$$
Since it passes through $A=(0,1)$,
$$AE:\quad y=1-\frac{x}{2}.$$
The line $OB$ is $x=1$. Therefore their intersection point is
$$K=\left(1,\frac12\right).$$
Finally,
$$O=(1,1),\qquad B=(1,0).$$
All three points $O$, $K$, and $B$ lie on the vertical line $x=1$, and
$$OK=1-\frac12=\frac12, \qquad KB=\frac12-0=\frac12.$$
Hence
$$OK=KB.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the determination of $O$. The tangents to the unit circle at $(0,1)$ and $(1,0)$ are $y=1$ and $x=1$ because the tangent at a point of the unit circle is perpendicular to the corresponding radius. Their intersection is indeed $(1,1)$.
The second delicate step is the computation of $E$. The line $OC$ has slope
$$\frac{-1-1}{0-1}=2,$$
hence equation $y=2x-1$. Substituting into the circle gives
$$x^2+(2x-1)^2=1,$$
which simplifies to
$$5x^2-4x=0.$$
The roots are $0$ and $4/5$. Since $x=0$ corresponds to $C=(0,-1)$, the second intersection must be $E=(4/5,3/5)$.
The third delicate step is locating $K$. Using the points $A=(0,1)$ and $E=(4/5,3/5)$, the slope of $AE$ is
$$\frac{3/5-1}{4/5}=-\frac12.$$
Thus $AE$ has equation $y=1-\frac{x}{2}$. Setting $x=1$ gives $y=1/2$. Since $O$ and $B$ have $y$-coordinates $1$ and $0$, the point $K=(1,1/2)$ is exactly their midpoint.
Alternative Approaches
A synthetic proof can be obtained by applying projective geometry on the complete quadrilateral formed by the tangents at $A$ and $B$, together with the secants $OC$ and $AE$. After expressing the relevant cross-ratios on the pencil through $O$, one shows that the point where $AE$ meets $OB$ divides $OB$ harmonically with the point at infinity of the direction $OB$. This forces $K$ to be the midpoint of $OB$.
Another approach uses affine coordinates adapted to the tangents. One introduces coordinates with origin at $O$ and axes along the sides of the angle. The circle then has equation
$$x^2+y^2-2ax-2ay+a^2=0.$$
The points $C$, $E$, and $K$ can be computed from the resulting equations. The calculations are longer than in the normalized unit-circle model, which is why the chosen proof is the most economical.