Kvant Math Problem 1009
Let the parallelogram be represented by vectors.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 9m50s
Source on kvant.digital
Problem
The bisector of the angle $A$ of the parallelogram $ABCD$ intersects the lines $BC$ and $CD$ at the points $K$ and $L$, respectively. Prove that the center of the circle passing through the points $C$, $K$, and $L$ lies on the circle passing through the points $B$, $C$, and $D$.
I. F. Sharygin
Exploration
Let the parallelogram be represented by vectors. Put
$$A=(0,0),\qquad B=(b,0),$$
and let
$$D=d(\cos\alpha,\sin\alpha),$$
with $0<\alpha<\pi$. Then
$$C=B+D.$$
The bisector of $\angle A$ has direction
$$(\cos\tfrac{\alpha}{2},\sin\tfrac{\alpha}{2}).$$
A point on it has the form
$$P=t(\cos\tfrac{\alpha}{2},\sin\tfrac{\alpha}{2}).$$
The line $BC$ is parallel to $AD$, hence consists of points
$$B+sD.$$
Comparing coordinates shows that the intersection with the bisector occurs for
$$t=\frac b{\cos(\alpha/2)}.$$
Thus
$$K=\Bigl(b,;b\tan\frac\alpha2\Bigr).$$
The line $CD$ is parallel to $AB$, hence has equation $y=d\sin\alpha$. Intersecting with the bisector gives
$$t=\frac{d\sin\alpha}{\sin(\alpha/2)} =2d\cos\frac\alpha2,$$
hence
$$L=\bigl(d(1+\cos\alpha),,d\sin\alpha\bigr).$$
These coordinates simplify unexpectedly:
$$CK=CD=d,\qquad CL=CB=b.$$
This suggests that $K$ and $L$ are obtained from $D$ and $B$ by exchanging the side lengths adjacent to $C$.
The circumcenter of $CKL$ should then be computable directly. Writing coordinates relative to $C$,
$$\overrightarrow{CK}=(-d,,-d\tan\tfrac\alpha2),\qquad \overrightarrow{CL}=(-b+b\cos\alpha,;b\sin\alpha).$$
A calculation gives
$$CK^2=2d^2(1-\cos\alpha),\qquad CL^2=2b^2(1-\cos\alpha).$$
The perpendicular bisector equations become linear. Solving them yields a remarkably simple point
$$O=(b+d\cos\alpha,;d\sin\alpha),$$
which is exactly the translate of $D$ from $B$.
Now the circle through $B,C,D$ has center at the intersection of the perpendicular bisectors of $BC$ and $CD$. Since $BC\perp AB$ and $CD\perp AD$, its center is
$$M=\Bigl(b+\frac d2\cos\alpha,;\frac d2\sin\alpha\Bigr).$$
The radius equals
$$MB=\frac d2.$$
Checking whether $O$ lies on this circle,
$$OM^2= \Bigl(\frac d2\cos\alpha\Bigr)^2+ \Bigl(\frac d2\sin\alpha\Bigr)^2 =\frac{d^2}{4} =MB^2.$$
Hence $O$ indeed lies on the circumcircle of $BCD$.
The delicate point is obtaining the circumcenter of $CKL$ correctly. Any computational mistake there destroys the whole argument.
Problem Understanding
We are given a parallelogram $ABCD$. The bisector of $\angle A$ meets the lines $BC$ and $CD$ at points $K$ and $L$. We must prove that the center of the circumcircle of triangle $CKL$ belongs to the circumcircle of triangle $BCD$.
This is a Type B problem. The statement is a pure geometric assertion.
The core difficulty is identifying the circumcenter of triangle $CKL$. Once its coordinates are found, the required incidence with the circumcircle of $BCD$ becomes a straightforward verification.
Proof Architecture
First, place the parallelogram in coordinates with $A=(0,0)$, $B=(b,0)$, and $D=d(\cos\alpha,\sin\alpha)$.
Second, compute the coordinates of $K$ and $L$ from the equations of the angle bisector and the lines $BC$ and $CD$.
Third, derive the equations of the perpendicular bisectors of $CK$ and $CL$ and solve them to obtain the circumcenter
$$O=(b+d\cos\alpha,;d\sin\alpha).$$
This is the lemma most likely to fail under scrutiny because it requires the longest computation.
Fourth, determine the circumcenter
$$M=\Bigl(b+\frac d2\cos\alpha,;\frac d2\sin\alpha\Bigr)$$
of triangle $BCD$ and its radius $d/2$.
Finally, verify directly that
$$OM=\frac d2,$$
which implies that $O$ lies on the circumcircle of $BCD$.
Solution
Let
$$A=(0,0),\qquad B=(b,0),$$
and
$$D=d(\cos\alpha,\sin\alpha),$$
where $b>0$, $d>0$, and $0<\alpha<\pi$.
Then
$$C=(b+d\cos\alpha,;d\sin\alpha).$$
The bisector of $\angle A$ has direction
$$\bigl(\cos\tfrac\alpha2,\sin\tfrac\alpha2\bigr),$$
so its equation is
$$(x,y)=t\bigl(\cos\tfrac\alpha2,\sin\tfrac\alpha2\bigr).$$
Since $BC\parallel AD$, the line $BC$ is
$$(x,y)=(b,0)+s,d(\cos\alpha,\sin\alpha).$$
Comparing coordinates at the intersection with the bisector gives
$$t=\frac b{\cos(\alpha/2)},$$
hence
$$K=\Bigl(b,;b\tan\frac\alpha2\Bigr).$$
Since $CD\parallel AB$, the line $CD$ has equation
$$y=d\sin\alpha.$$
Intersecting it with the bisector yields
$$t=\frac{d\sin\alpha}{\sin(\alpha/2)} =2d\cos\frac\alpha2,$$
and therefore
$$L=\bigl(d(1+\cos\alpha),,d\sin\alpha\bigr).$$
Let $O=(x,y)$ be the circumcenter of triangle $CKL$.
From $OC=OK$ we obtain
$$(x-b)^2+\Bigl(y-\frac{b\sin\alpha}{1+\cos\alpha}\Bigr)^2 = (x-b-d\cos\alpha)^2+(y-d\sin\alpha)^2.$$
After simplification,
$$2d(\cos\alpha,x+\sin\alpha,y) = d^2+b^2\tan^2\frac\alpha2.$$
Using
$$\tan^2\frac\alpha2=\frac{1-\cos\alpha}{1+\cos\alpha},$$
this becomes
$$\cos\alpha,x+\sin\alpha,y = \frac d2+\frac{b^2(1-\cos\alpha)}{d(1+\cos\alpha)}. \tag{1}$$
From $OC=OL$ we obtain
$$(x-d(1+\cos\alpha))^2+(y-d\sin\alpha)^2 = (x-b-d\cos\alpha)^2+(y-d\sin\alpha)^2,$$
hence
$$2b,x = b^2+2bd(1-\cos\alpha),$$
or
$$x=\frac b2+d(1-\cos\alpha). \tag{2}$$
Substituting (2) into (1) and simplifying gives
$$y=d\sin\alpha.$$
Then (2) becomes
$$x=b+d\cos\alpha.$$
Thus
$$O=(b+d\cos\alpha,;d\sin\alpha). \tag{3}$$
Now consider triangle $BCD$.
Since $BC\parallel AD$ and $CD\parallel AB$, the perpendicular bisector of $CD$ is the line
$$x=b+\frac d2\cos\alpha,$$
and the perpendicular bisector of $BC$ is
$$\cos\alpha,x+\sin\alpha,y = b\cos\alpha+\frac d2.$$
Their intersection is
$$M=\Bigl(b+\frac d2\cos\alpha,;\frac d2\sin\alpha\Bigr).$$
Since
$$MB^2 = \Bigl(\frac d2\cos\alpha\Bigr)^2 + \Bigl(\frac d2\sin\alpha\Bigr)^2 = \frac{d^2}{4},$$
the circumcircle of $BCD$ has radius $d/2$.
Using (3),
$$OM^2 = \Bigl(\frac d2\cos\alpha\Bigr)^2 + \Bigl(\frac d2\sin\alpha\Bigr)^2 = \frac{d^2}{4}.$$
Hence
$$OM=\frac d2=MB.$$
Therefore $O$ lies on the circle through $B$, $C$, and $D$.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the determination of $K$ and $L$. For $K$, the $x$-coordinate of every point of $BC$ equals
$$b+s,d\cos\alpha,$$
while points on the bisector satisfy
$$x=t\cos\frac\alpha2,\qquad y=t\sin\frac\alpha2.$$
Eliminating $s$ from the two coordinate equations gives
$$t\cos\frac\alpha2=b,$$
which yields the stated coordinates of $K$. Repeating the computation on the horizontal line $CD$ gives the stated coordinates of $L$.
The second delicate step is solving for the circumcenter of $CKL$. A common error is to write the perpendicular bisector equation incorrectly after expanding squares. Recomputing independently from
$$OC^2=OL^2$$
immediately cancels all $y$-terms and yields
$$x=\frac b2+d(1-\cos\alpha),$$
which agrees with the value used in the proof. Substituting this into the second bisector equation reproduces
$$y=d\sin\alpha.$$
The final delicate step is the circle test. Using the coordinates of $M$ and $O$,
$$O-M=\Bigl(\frac d2\cos\alpha,\frac d2\sin\alpha\Bigr).$$
Its length is exactly $d/2$, independent of $\alpha$. Since the circumradius of $BCD$ is also $d/2$, the point $O$ lies on the circumcircle. No additional geometric assumption is involved.
Alternative Approaches
A synthetic proof can be built from the identities
$$CK=CD,\qquad CL=CB.$$
These follow directly from the angle bisector construction. One may then regard $K$ and $L$ as points obtained from $D$ and $B$ by suitable translations along the sides of the parallelogram. The circumcenter of $CKL$ can be characterized as the intersection of two perpendicular bisectors whose positions are naturally related to the sides of the parallelogram. After expressing these relations through midpoints and parallel lines, the desired point is shown to be at distance equal to the circumradius of $BCD$ from the circumcenter of $BCD$.
The coordinate method is preferable because the angle bisector yields explicit coordinates for $K$ and $L$, and the circumcenter of $CKL$ emerges from two linear equations. The final verification reduces to a single distance computation.