Kvant Math Problem 41
Let the circle have center $O$ and radius $R$.
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Solve time: 6m22s
Source on kvant.digital
Problem
Given a circle, its diameter $AB$, and a point $C$ on this diameter. Construct on the circle two points $X$ and $Y$, symmetric with respect to the diameter $AB$, such that the line $YC$ is perpendicular to the line $XA$.
All-Union Mathematical Olympiad (1970, Grade 8)
Exploration
Let the circle have center $O$ and radius $R$. Put the diameter $AB$ on the $x$-axis, with
$$A=(-R,0), \qquad B=(R,0),$$
and let
$$C=(c,0),$$
where $-R\le c\le R$.
Since $X$ and $Y$ are symmetric with respect to the diameter $AB$, they must have coordinates
$$X=(u,v), \qquad Y=(u,-v),$$
with
$$u^2+v^2=R^2.$$
The condition $YC\perp XA$ translates into the product of slopes being $-1$.
The slope of $XA$ is
$$\frac{v}{u+R},$$
and the slope of $YC$ is
$$\frac{-v}{u-c}.$$
Hence
$$\frac{v}{u+R}\cdot \frac{-v}{u-c}=-1.$$
Since $v^2=R^2-u^2$, this becomes
$$-(R^2-u^2)=-(u+R)(u-c).$$
Therefore
$$R^2-u^2=(u+R)(u-c).$$
Expanding,
$$R^2-u^2=u^2+(R-c)u-Rc,$$
hence
$$2u^2+(R-c)u-R(R+c)=0.$$
Trying $u=R$ gives
$$2R^2+R(R-c)-R(R+c)=2R^2\ne0.$$
Trying $u=-R$ gives $0$, so $u=-R$ is a root. Factoring,
$$2u^2+(R-c)u-R(R+c) =(u+R)(2u-R-c).$$
The root $u=-R$ corresponds to the endpoint $A$ itself, for which $X=Y=A$, a degenerate configuration. The nontrivial root is
$$u=\frac{R+c}{2}.$$
The crucial point is to show that this value always determines points of the circle. Since
$$-R\le c\le R,$$
we have
$$0\le R+c\le 2R,$$
and therefore
$$-R\le \frac{R+c}{2}\le R.$$
Thus the vertical line
$$x=\frac{R+c}{2}$$
meets the circle in two symmetric points, giving the required construction.
The coordinate computation suggests a very simple geometric interpretation: the common abscissa of $X$ and $Y$ is the midpoint of the segment $CB$.
Problem Understanding
We are given a circle, a diameter $AB$, and a point $C$ on this diameter. We must construct two points $X$ and $Y$ on the circle, symmetric with respect to the diameter $AB$, such that the line through $Y$ and $C$ is perpendicular to the line through $X$ and $A$.
This is a Type D problem, an existence and construction problem.
The core difficulty is to translate the perpendicularity condition into a geometric condition determining where the symmetric pair $X,Y$ must lie on the circle.
The answer is: let $M$ be the midpoint of $CB$. Draw the line through $M$ perpendicular to $AB$. Its intersections with the circle are the required points $X$ and $Y$. The exploratory calculation indicates that the common projection of $X$ and $Y$ onto $AB$ is exactly the midpoint of $CB$.
Proof Architecture
First, introduce coordinates with the diameter $AB$ as the $x$-axis and represent the symmetric points as $(u,\pm v)$.
Second, translate the condition $YC\perp XA$ into an algebraic equation relating $u$ and $v$.
Third, use the circle equation $u^2+v^2=R^2$ to eliminate $v$ and obtain a quadratic equation for $u$.
Fourth, factor this quadratic and show that the nondegenerate solution is $u=(R+c)/2$.
Fifth, identify geometrically that $(R+c)/2$ is the abscissa of the midpoint of $CB$.
Sixth, prove that the intersections of the circle with the perpendicular through that midpoint satisfy the required perpendicularity condition.
The lemma most likely to fail under scrutiny is the derivation of the quadratic equation for $u$, since any algebraic error there would invalidate the construction.
Solution
Let the circle have center $O$ and radius $R$. Choose coordinates so that
$$A=(-R,0), \qquad B=(R,0),$$
and let
$$C=(c,0),$$
with $-R\le c\le R$.
Suppose $X$ and $Y$ are points of the circle symmetric with respect to the diameter $AB$. Then
$$X=(u,v), \qquad Y=(u,-v),$$
and
$$u^2+v^2=R^2.$$
The condition $YC\perp XA$ means that the product of their slopes equals $-1$.
The slope of $XA$ is
$$\frac{v}{u+R},$$
and the slope of $YC$ is
$$\frac{-v}{u-c}.$$
Hence
$$\frac{v}{u+R}\cdot\frac{-v}{u-c}=-1.$$
Multiplying by $(u+R)(u-c)$ gives
$$v^2=(u+R)(u-c).$$
Since $u^2+v^2=R^2$,
$$R^2-u^2=(u+R)(u-c).$$
After expansion,
$$R^2-u^2=u^2+(R-c)u-Rc,$$
so
$$2u^2+(R-c)u-R(R+c)=0.$$
Substituting $u=-R$ shows that $u+R$ is a factor. Factoring,
$$2u^2+(R-c)u-R(R+c) =(u+R)(2u-R-c).$$
Therefore
$$u=-R \quad\text{or}\quad u=\frac{R+c}{2}.$$
The solution $u=-R$ corresponds to the degenerate point $A$. The nondegenerate solution is
$$u=\frac{R+c}{2}.$$
Let $M$ be the midpoint of the segment $CB$. Since
$$C=(c,0), \qquad B=(R,0),$$
the abscissa of $M$ is
$$\frac{c+R}{2}.$$
Thus every nondegenerate solution must satisfy that $X$ and $Y$ lie on the line through $M$ perpendicular to $AB$.
This yields the construction:
Construct the midpoint $M$ of $CB$. Draw the line through $M$ perpendicular to $AB$. Let its intersections with the circle be $X$ and $Y$.
It remains to verify that these points satisfy the required condition.
For these points,
$$u=\frac{R+c}{2}.$$
Repeating the computation above in reverse,
$$(u+R)(u-c) = R^2-u^2 = v^2.$$
Hence
$$\frac{v}{u+R}\cdot\frac{-v}{u-c} = -\frac{v^2}{(u+R)(u-c)} = -1.$$
Therefore the slopes of $XA$ and $YC$ have product $-1$, so
$$YC\perp XA.$$
The constructed points satisfy all requirements, and the construction is complete.
$$\boxed{\text{Take }M\text{ as the midpoint of }CB,\ \text{draw through }M\text{ the perpendicular to }AB;\ \text{its intersections with the circle are }X,Y.}$$
Verification of Key Steps
The first delicate step is the translation of perpendicularity into
$$v^2=(u+R)(u-c).$$
Starting directly from the coordinates,
$$m_{XA}=\frac{v}{u+R}, \qquad m_{YC}=\frac{-v}{u-c}.$$
Perpendicularity gives
$$m_{XA}m_{YC}=-1,$$
hence
$$-\frac{v^2}{(u+R)(u-c)}=-1,$$
which is equivalent to the stated identity. No hidden assumption is used except that the lines are not vertical. If one is vertical, the same equation is obtained by continuity and corresponds to the same constructed points.
The second delicate step is the factorization
$$2u^2+(R-c)u-R(R+c) =(u+R)(2u-R-c).$$
Expanding the right-hand side gives
$$2u^2-(R+c)u+2Ru-R(R+c) = 2u^2+(R-c)u-R(R+c),$$
which matches exactly.
The third delicate step is identifying the geometric meaning of
$$u=\frac{R+c}{2}.$$
Since $C=(c,0)$ and $B=(R,0)$, the midpoint of $CB$ has coordinate
$$\left(\frac{R+c}{2},0\right).$$
Thus the common projection of $X$ and $Y$ onto the diameter is precisely that midpoint, yielding the construction directly.
Alternative Approaches
A synthetic solution can be obtained by introducing the midpoint $M$ of $CB$ from the outset and letting the perpendicular through $M$ meet the circle at $X$ and $Y$. Since $AB$ is a diameter, the angles $AXB$ and $AYB$ are right angles. Applying the power of a point to $M$ gives
$$MX\cdot MY=MC\cdot MB.$$
Because $MX=MY$, one gets
$$MX^2=MC\cdot MB.$$
From the right triangles determined by the diameter, this identity can be transformed into a relation between the slopes of $XA$ and $YC$, yielding their perpendicularity.
The coordinate method is preferable here because the symmetry with respect to the diameter is represented immediately by $(u,\pm v)$, and the perpendicularity condition reduces to a single quadratic equation whose nondegenerate root reveals the construction in a transparent way.