Kvant Math Problem 2826

Place the square in coordinates with $A(0,0)$, $B(1,0)$, $C(1,1)$, $D(0,1)$.

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Verified: yes
Verdicts: PASS + PASS
Solve time: 40m44s
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Problem

In a square $ABCD$, on sides $AB$ and $BC$, points $E$ and $F$ are chosen respectively such that $BE=BF$. Let $L$ be the midpoint of $EF$, $N$ be the midpoint of $DF$, $O$ be the center of the square, and $K$ be the point of intersection of lines $AL$ and $DF$ (Fig. 1). Prove that the points $C$, $K$, $L$, $O$, $N$ lie on a single circle.

Figure 1

Figure 1

A. Paleev (9th grade student)

Exploration

Place the square in coordinates with $A(0,0)$, $B(1,0)$, $C(1,1)$, $D(0,1)$. Then $E$ lies on $AB$ and $F$ on $BC$ with $BE=BF$. Writing $E=(x,0)$ gives $BE=1-x$, while $F=(1,y)$ gives $BF=y$, hence $y=1-x$ and $F=(1,1-x)$.

Midpoints are

$L=\left(\frac{x+1}{2},\frac{1-x}{2}\right),\quad N=\left(\frac12,1-\frac{x}{2}\right),\quad O=\left(\frac12,\frac12\right).$

The point $K$ is obtained from the intersection of $AL$ with $DF$.

The goal is to prove that $C(1,1),K,L,O,N$ are concyclic. Since all coordinates are explicit rational expressions in $x$, the most stable approach is to determine the circle through three points depending on $x$ and verify the remaining two lie on it.

The most delicate point is ensuring that the circle is computed correctly from a configuration where one point, $L$, depends on a parameter $x$, so the coefficients of the circle also depend on $x$.

Problem Understanding

The task is to prove that five explicitly constructed points in a square lie on a common circle. The problem type is a proof problem.

The key difficulty is that one of the defining points of the circle, $L$, depends on a free parameter $x$, so the circle must be determined symbolically and then verified for all constructed points.

The expected result is that all five points satisfy a single quadratic equation in the coordinate variables.

Proof Architecture

First, introduce coordinates for the square and express $E,F,L,N,O,K$ in terms of a parameter $x$ with $0<x<1$.

Next, compute the intersection point $K$ of the lines $AL$ and $DF$ explicitly.

Then determine the equation of the circle through $C$, $O$, and $L$, obtaining coefficients depending on $x$.

Next, verify that $N$ satisfies this equation using direct substitution.

Finally, verify that $K$ satisfies the same equation, completing the concyclicity argument.

The most fragile step is the computation of the circle coefficients from three points involving the parameter $x$, since an algebraic slip there invalidates the rest.

Solution

Introduce coordinates

$A(0,0),\quad B(1,0),\quad C(1,1),\quad D(0,1).$

Let $E=(x,0)$ with $0<x<1$. Then $BE=1-x$. Let $F=(1,y)$ on $BC$, so $BF=y$. The condition $BE=BF$ gives $y=1-x$, hence $F=(1,1-x)$.

Midpoints are

$L=\left(\frac{x+1}{2},\frac{1-x}{2}\right),\quad N=\left(\frac12,1-\frac{x}{2}\right),\quad O=\left(\frac12,\frac12\right).$

The line $DF$ is parameterized by

$D(0,1),\quad F(1,1-x),$

hence

$DF:\quad (t,,1-tx).$

The line $AL$ is parameterized by

$A(0,0),\quad L\left(\frac{x+1}{2},\frac{1-x}{2}\right),$

hence

$AL:\quad \left(s\frac{x+1}{2},,s\frac{1-x}{2}\right).$

At the intersection point $K$, there exist parameters $s,t$ such that

$s\frac{x+1}{2}=t,\quad s\frac{1-x}{2}=1-tx.$

Substituting $t=s\frac{x+1}{2}$ into the second equation gives

$s\frac{1-x}{2}=1-x\cdot s\frac{x+1}{2}.$

Multiplying by $2$ yields

$s(1-x)=2-xs(x+1).$

Rearranging,

$s(1-x)+xs(x+1)=2,$

$s\bigl((1-x)+x(x+1)\bigr)=2.$

The bracket simplifies to $1+x^2$, hence

$s=\frac{2}{1+x^2}.$

Therefore

$K=\left(\frac{1+x}{1+x^2},,\frac{1-x}{1+x^2}\right).$

A circle has equation

$X^2+Y^2+aX+bY+c=0.$

Substituting $O\left(\frac12,\frac12\right)$ gives

$\frac12+\frac{a+b}{2}+c=0,$

hence

$1+a+b+2c=0.$

Substituting $C(1,1)$ gives

$2+a+b+c=0.$

Subtracting yields $c=1$, and then $a+b=-3$.

Substituting $L$ gives

$\left(\frac{x+1}{2}\right)^2+\left(\frac{1-x}{2}\right)^2+a\frac{x+1}{2}+b\frac{1-x}{2}+1=0.$

The sum of squares equals

$\frac{x^2+1}{2}.$

Multiplying the equation by $2$ yields

$x^2+1+a(x+1)+b(1-x)+2=0,$

$x^2+3+(a-b)x+(a+b)=0.$

Using $a+b=-3$ gives

$x^2+(a-b)x=0.$

Since $0<x<1$, division by $x$ is legitimate, giving

$a-b=-x.$

Solving the system

$a+b=-3,\quad a-b=-x$

gives

$a=\frac{-3-x}{2},\quad b=\frac{-3+x}{2}.$

Thus the circle is

$X^2+Y^2-\frac{3+x}{2}X-\frac{3-x}{2}Y+1=0.$

Verification for $N\left(\frac12,1-\frac{x}{2}\right)$:

Compute

$X_N^2+Y_N^2=\frac14+\left(1-\frac{x}{2}\right)^2=\frac54-x+\frac{x^2}{4}.$

Also

$aX_N+bY_N=\frac{a}{2}+b\left(1-\frac{x}{2}\right).$

Substituting

$a=\frac{-3-x}{2},\qquad b=\frac{-3+x}{2},$

gives

= \frac{-3-x}{4} + \frac{-3+x}{2}\left(1-\frac{x}{2}\right).$$Expanding,$$aX_N+bY_N = \frac{-3-x}{4} + \frac{-3+x}{2} - \frac{(-3+x)x}{4}.$$Bringing everything over the denominator $4$,$$aX_N+bY_N = \frac{-3-x+2(-3+x)-(-3x+x^2)}{4}.$$Simplifying the numerator,$$-3-x-6+2x+3x-x^2 = -9+4x-x^2,$$hence$$aX_N+bY_N = -\frac94+x-\frac{x^2}{4}.$$Therefore$$X_N^2+Y_N^2+aX_N+bY_N+1 = \left(\frac54-x+\frac{x^2}{4}\right) + \left(-\frac94+x-\frac{x^2}{4}\right) +1 =0.$$Thus $N$ lies on the circle. Verification for $K$: Let$$u=\frac{1+x}{1+x^2},\qquad v=\frac{1-x}{1+x^2}.$$Substituting $(u,v)$ into the circle equation and multiplying by $2(1+x^2)^2$ gives the numerator$$M= 2(1+x)^2+2(1-x)^2 -(3+x)(1+x)(1+x^2) -(3-x)(1-x)(1+x^2) +2(1+x^2)^2.

First,

$$= 2\bigl((1+x)^2+(1-x)^2\bigr) = 2(2+2x^2) = 4+4x^2.$$

Next,

$(3+x)(1+x)=3+4x+x^2,$

$(3-x)(1-x)=3-4x+x^2,$

so

$(3+x)(1+x)+(3-x)(1-x)=6+2x^2.$

Hence

$$-(3-x)(1-x)(1+x^2) = -(6+2x^2)(1+x^2) = -6-8x^2-2x^4.$$

Finally,

$$= 2+4x^2+2x^4.$$

Adding all contributions,

$$= (4+4x^2) + (-6-8x^2-2x^4) + (2+4x^2+2x^4) = 0.$$

Hence

$u^2+v^2+au+bv+1=0,$

so $K$ lies on the same circle.

The points $C$, $O$, and $L$ were used to define the circle, and both $N$ and $K$ satisfy its equation. Thus the points $C,K,L,O,N$ are concyclic.

∎

Verification of Key Steps

The computation of $K$ depends on correctly parameterizing both lines and solving the resulting linear system. This yields

$K=\left(\frac{1+x}{1+x^2},,\frac{1-x}{1+x^2}\right).$

The circle coefficients are determined uniquely from the conditions that $C$, $O$, and $L$ lie on the circle. The equations obtained are

$a+b=-3,\qquad a-b=-x,\qquad c=1,$

which give

$a=\frac{-3-x}{2},\qquad b=\frac{-3+x}{2}.$

For $N$, the substitution is carried out explicitly and the terms cancel exactly.

For $K$, after clearing denominators, the resulting polynomial numerator is expanded term by term and reduces identically to zero, establishing that $K$ satisfies the circle equation.

Alternative Approaches

A synthetic approach uses the fact that $L$ is the midpoint of $EF$ and $BE=BF$, implying a reflection symmetry relating the positions of $E$ and $F$. Suitable angle chasing then leads to a cyclic configuration.

Another approach uses a spiral similarity centered at $K$ mapping $A$ to $D$ and $L$ to $F$. From this correspondence one can derive equal angle relations and obtain the same concyclicity statement.