Kvant Math Problem 633
Let the circle have center $O$ and radius $R$.
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Problem
On a diameter $AC$ of a certain circle, a point $E$ is given. Draw through it a chord $BD$ so that the area of the quadrilateral $ABCD$ is maximized.
I. F. Sharygin
All-Union Mathematical Olympiad (XIV, 1980, 9th grade)
Exploration
Let the circle have center $O$ and radius $R$. The diameter is $AC$, and the point $E$ lies on $AC$. A chord $BD$ passes through $E$.
The area to be maximized is the area of quadrilateral $ABCD$. Since $AC$ is a diagonal of this quadrilateral, its area equals
$$[ABCD]=[ABC]+[ACD].$$
Because $AC$ is a diameter, $\angle ABC=\angle ADC=90^\circ$. Hence
$$[ABC]=\frac12,AB\cdot BC,\qquad [ACD]=\frac12,AD\cdot DC.$$
This expression is not immediately convenient because the condition involves the chord $BD$ passing through a fixed point $E$.
A better idea is to use the common diagonal $AC$. The area of $ABCD$ equals
$$\frac12,AC,(h_B+h_D),$$
where $h_B,h_D$ are the distances from $B,D$ to the line $AC$.
Since $B$ and $D$ lie on opposite sides of $AC$, $h_B+h_D$ is exactly the length of the projection of the chord $BD$ onto the direction perpendicular to $AC$.
Introduce coordinates with
$$O=(0,0),\qquad A=(-R,0),\qquad C=(R,0),$$
and let
$$E=(e,0).$$
Let the chord through $E$ make an angle $\theta$ with $AC$. Parametrize the line by
$$(x,y)=(e,0)+t(\cos\theta,\sin\theta).$$
Its intersections with the circle $x^2+y^2=R^2$ correspond to values $t_1,t_2$ satisfying
$$(e+t\cos\theta)^2+t^2\sin^2\theta=R^2,$$
that is,
$$t^2+2e\cos\theta,t+(e^2-R^2)=0.$$
Hence
$$t_2-t_1 =2\sqrt{R^2-e^2\sin^2\theta}.$$
The projection of $BD$ onto the perpendicular to $AC$ has length
$$(t_2-t_1)|\sin\theta| = 2|\sin\theta|\sqrt{R^2-e^2\sin^2\theta}.$$
Therefore
$$[ABCD] = 2R,|\sin\theta|\sqrt{R^2-e^2\sin^2\theta}.$$
Set
$$u=\sin^2\theta.$$
Then maximizing the area is equivalent to maximizing
$$f(u)=u(R^2-e^2u),\qquad 0\le u\le1.$$
The crucial point is determining the maximum of this quadratic and translating it back into a geometric condition on the chord.
Problem Understanding
A circle with diameter $AC$ is given, together with a fixed point $E$ on the diameter. Among all chords $BD$ passing through $E$, we must determine which one makes the area of the cyclic quadrilateral $ABCD$ as large as possible.
This is a Type C problem. We must determine the maximizing chord and prove that no other chord yields a larger area.
The core difficulty is expressing the area of $ABCD$ in terms of the direction of the chord through $E$, then optimizing that expression under the constraint that the chord passes through the fixed interior point $E$.
The answer will be that the optimal chord is the one whose distance from the center satisfies a specific relation. Equivalently, if $OE=e$, then the chord should make an angle $\theta$ with the diameter such that
$$\sin^2\theta= \min!\left(1,\frac{R^2}{2e^2}\right).$$
When $e\le \dfrac{R}{\sqrt2}$, the maximizing chord is perpendicular to the diameter. When $e>\dfrac{R}{\sqrt2}$, there are two symmetric maximizing chords.
Proof Architecture
The area of $ABCD$ equals $\dfrac12,AC,(h_B+h_D)$, because $AC$ is a diagonal splitting the quadrilateral into two triangles.
For a chord through $E$, the quantity $h_B+h_D$ equals the projection of $BD$ onto the direction perpendicular to $AC$, because $B$ and $D$ lie on opposite sides of $AC$.
If the chord through $E=(e,0)$ makes angle $\theta$ with $AC$, then its intersections with the circle occur at parameters whose difference is $2\sqrt{R^2-e^2\sin^2\theta}$, obtained from the quadratic equation describing the intersections.
Hence
$$[ABCD] = 2R,|\sin\theta|\sqrt{R^2-e^2\sin^2\theta}.$$
After setting $u=\sin^2\theta$, maximizing the area reduces to maximizing the quadratic function
$$u(R^2-e^2u)$$
on the interval $[0,1]$.
The hardest step is converting the geometric condition into the analytic formula for the area and then carrying out the optimization without overlooking the boundary case $e\le R/\sqrt2$.
Solution
Let the circle have center $O$ and radius $R$. Choose coordinates so that
$$A=(-R,0),\qquad C=(R,0),$$
and let
$$E=(e,0),$$
where $e=OE$.
For any chord $BD$ through $E$, the diagonal $AC$ divides the quadrilateral $ABCD$ into triangles $ABC$ and $ACD$. Hence
$$[ABCD]=[ABC]+[ACD].$$
If $h_B$ and $h_D$ denote the distances from $B$ and $D$ to the line $AC$, then
$$[ABC]=\frac12,AC,h_B, \qquad [ACD]=\frac12,AC,h_D.$$
Since $AC=2R$,
$$[ABCD] = R(h_B+h_D).$$
The points $B$ and $D$ lie on opposite sides of the diameter $AC$. Therefore $h_B+h_D$ is the length of the projection of the chord $BD$ onto the axis perpendicular to $AC$.
Let the chord make an angle $\theta$ with $AC$. The line containing it has equation
$$(x,y)=(e,0)+t(\cos\theta,\sin\theta).$$
Substituting into the circle equation $x^2+y^2=R^2$ gives
$$(e+t\cos\theta)^2+t^2\sin^2\theta=R^2,$$
or
$$t^2+2e\cos\theta,t+(e^2-R^2)=0.$$
Let $t_1,t_2$ be the roots. Their difference is
$$t_2-t_1 = \sqrt{(2e\cos\theta)^2-4(e^2-R^2)} = 2\sqrt{R^2-e^2\sin^2\theta}.$$
Thus the chord length is
$$BD=2\sqrt{R^2-e^2\sin^2\theta}.$$
Its projection onto the perpendicular to $AC$ has length
$$BD,|\sin\theta| = 2|\sin\theta|\sqrt{R^2-e^2\sin^2\theta}.$$
Consequently,
$$[ABCD] = 2R,|\sin\theta|\sqrt{R^2-e^2\sin^2\theta}.$$
Since the area depends only on $\sin^2\theta$, put
$$u=\sin^2\theta,\qquad 0\le u\le1.$$
Maximizing the area is equivalent to maximizing its square:
$$[ABCD]^2 = 4R^2,u(R^2-e^2u).$$
Define
$$f(u)=u(R^2-e^2u).$$
This is a concave quadratic. Its derivative is
$$f'(u)=R^2-2e^2u.$$
Hence the unconstrained maximum occurs at
$$u_0=\frac{R^2}{2e^2}.$$
Two cases arise.
If
$$e\le\frac{R}{\sqrt2},$$
then $u_0\ge1$. Since $f$ is increasing on $[0,1]$, its maximum on that interval is attained at
$$u=1.$$
Therefore
$$\sin^2\theta=1,$$
which means
$$\theta=\frac{\pi}{2}.$$
The maximizing chord is perpendicular to the diameter $AC$.
If
$$e>\frac{R}{\sqrt2},$$
then $0<u_0<1$, and the maximum is attained at
$$u=u_0=\frac{R^2}{2e^2}.$$
Thus the maximizing chord is characterized by
$$\sin^2\theta=\frac{R^2}{2e^2}.$$
There are two such chords, symmetric with respect to the diameter $AC$.
The maximal area equals
$$2R\sqrt{f(u_{\max})}.$$
For $e\le R/\sqrt2$,
$$[ABCD]_{\max} = 2R\sqrt{R^2-e^2}.$$
For $e>R/\sqrt2$,
$$[ABCD]_{\max} = \frac{R^3}{e}.$$
Hence the required chord is:
$$\boxed{\sin^2\angle(BD,AC)= \min!\left(1,\frac{R^2}{2,OE^2}\right)}.$$
In particular, if $OE\le R/\sqrt2$, the optimal chord is perpendicular to the diameter $AC$; otherwise it is one of the two symmetric chords satisfying
$$\sin^2\angle(BD,AC)=\frac{R^2}{2,OE^2}.$$
Verification of Key Steps
The first delicate point is the area formula. Since $AC$ is a diagonal,
$$[ABCD]=[ABC]+[ACD].$$
Each triangle has base $AC$, and their heights are the perpendicular distances from $B$ and $D$ to $AC$. Because $B$ and $D$ lie on opposite sides of the diameter, the heights add rather than subtract. This yields
$$[ABCD]=R(h_B+h_D).$$
The second delicate point is the chord length computation. Substituting the parametric equation of the line through $E$ into the circle equation produces a quadratic in $t$. The distance between the two intersection parameters equals the square root of the discriminant. Computing the discriminant gives
$$4(R^2-e^2\sin^2\theta),$$
hence
$$BD=2\sqrt{R^2-e^2\sin^2\theta}.$$
Any sign error in expanding the quadratic would change the optimization problem completely.
The third delicate point is the boundary between the two cases. The critical value of
$$u_0=\frac{R^2}{2e^2}$$
lies inside the interval $[0,1]$ exactly when
$$e\ge\frac{R}{\sqrt2}.$$
If this condition is overlooked, one would incorrectly claim that the stationary point always gives the maximum. For example, when $e=\frac R2$, the formula gives $u_0=2$, which is impossible; the true maximum then occurs at the endpoint $u=1$, corresponding to the perpendicular chord.
Alternative Approaches
A synthetic approach is possible by expressing the area as
$$[ABCD]=[ABC]+[ACD] =\frac12(AB\cdot BC+AD\cdot DC).$$
Using the right triangles subtended by the diameter, one may rewrite these products through the coordinates of $B$ and $D$, then use the condition that $BD$ passes through the fixed point $E$. After introducing a suitable parameter for the direction of the chord, the same quadratic optimization emerges.
The coordinate method used above is preferable because the dependence on the direction of the chord becomes explicit immediately. The area reduces to a single-variable quadratic maximization, and the boundary case $OE\le R/\sqrt2$ appears naturally from the interval restriction $0\le\sin^2\theta\le1$.