Kvant Math Problem 2844

We place square $ABCD$ in the coordinate plane with $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, and $D=(0,1)$.

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Verified: no
Verdicts: FAIL + UNKNOWN
Solve time: 33m26s
Source on kvant.digital

Problem

The hostess took a piece of meat out of the refrigerator, and kittens gathered around her. Once every minute, the hostess cuts off a piece of meat and feeds it to one of the kittens (at her choice), with each piece having to be the same fraction of the piece from which it is cut. After some time, the hostess puts the remaining meat back into the refrigerator. Can she feed the kittens equally if there are

  1. two kittens;
  2. three kittens;
  3. four kittens?

A. Kushnir, V. Retinsky

Cities of the Tour (XLIV)

Exploration

We place square $ABCD$ in the coordinate plane with $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, and $D=(0,1)$. Let $E$ be a point on $AB$ and $F$ a point on $BC$ such that $BE=BF$. Parametrize $E$ as $E=(x,0)$ with $0\le x\le 1$, giving $BE=1-x$. Then $BF=y$, so $y=1-x$ and $F=(1,1-x)$. Denote $O$ as the center of the square, $O=\left(\frac12,\frac12\right)$. Define the midpoint $L$ of segment $EF$ as

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

Let $N$ be the midpoint of segment $AD$, giving $N=\left(\frac12,1-\frac{x}{2}\right)$.

We define point $K$ as the intersection of lines $AL$ and $DF$. Explicit computation of $K$ is necessary to verify concyclicity. Line $AL$ passes through points $A=(0,0)$ and $L=\left(\frac{x+1}{2}, \frac{1-x}{2}\right)$, so its parametric form is

$X=\lambda \frac{x+1}{2}, \quad Y=\lambda \frac{1-x}{2}.$

Line $DF$ passes through points $D=(0,1)$ and $F=(1,1-x)$, so its parametric form is

$X=\mu, \quad Y=1-\mu (1-x).$

Setting the parametric forms equal, we solve for $\lambda$ and $\mu$:

$\lambda \frac{x+1}{2} = \mu, \quad \lambda \frac{1-x}{2} = 1 - \mu (1-x).$

From the first equation, $\mu = \lambda \frac{x+1}{2}$. Substituting into the second equation gives

$\lambda \frac{1-x}{2} = 1 - \lambda \frac{x+1}{2} (1-x) = 1 - \lambda \frac{(x+1)(1-x)}{2} = 1 - \lambda \frac{1-x^2}{2}.$

Simplifying,

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

Thus $\lambda = \frac{2}{2 - x - x^2}$. Then $\mu = \frac{x+1}{2} \cdot \frac{2}{2 - x - x^2} = \frac{2(x+1)}{2(2 - x - x^2)} = \frac{x+1}{2 - x - x^2}$. The coordinates of $K$ are therefore

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

This explicit formula allows us to verify concyclicity of $C, K, L, O, N$.

Problem Understanding

The problem is to prove that five points in a square lie on a common circle for arbitrary position of $E$ along side $AB$, equivalently arbitrary $x\in[0,1]$. The points are $C=(1,1)$, $K=\left(\frac{x+1}{2 - x - x^2}, \frac{1-x}{2 - x - x^2}\right)$, $L=\left(\frac{x+1}{2}, \frac{1-x}{2}\right)$, $O=\left(\frac12,\frac12\right)$, and $N=\left(\frac12,1-\frac{x}{2}\right)$. A complete proof requires either an algebraic verification that all five points satisfy the same circle equation or a geometric argument that guarantees concyclicity independent of $x$.

Solution

To prove concyclicity algebraically, we write the general equation of a circle as

$(X - h)^2 + (Y - k)^2 = r^2.$

We can determine $(h,k)$ and $r$ by solving the system using points $C$, $O$, and $L$. Let $C=(1,1)$, $O=(\frac12,\frac12)$, $L=(\frac{x+1}{2},\frac{1-x}{2})$. The center $(h,k)$ satisfies

$(1 - h)^2 + (1 - k)^2 = r^2,$

$\left(\frac12 - h\right)^2 + \left(\frac12 - k\right)^2 = r^2,$

$\left(\frac{x+1}{2} - h\right)^2 + \left(\frac{1-x}{2} - k\right)^2 = r^2.$

Subtracting the second equation from the first gives

$(1-h)^2 + (1-k)^2 - \left(\frac12 - h\right)^2 - \left(\frac12 - k\right)^2 = 0,$

which simplifies to

$(1 - h)^2 - \left(\frac12 - h\right)^2 + (1 - k)^2 - \left(\frac12 - k\right)^2 = 0.$

Compute each term: $(1 - h)^2 - (\frac12 - h)^2 = (1 - h - (\frac12 - h)) (1 - h + \frac12 - h) = (\frac12)(\frac32 - 2h) = \frac34 - h.$ Wait, compute carefully.

Compute $(1 - h)^2 - (\frac12 - h)^2 = (1 - h)^2 - (\frac12 - h)^2 = (1 - 2h + h^2) - (\frac14 - h + h^2) = 1 - 2h + h^2 - \frac14 + h - h^2 = 3/4 - h.$ Similarly, $(1 - k)^2 - (\frac12 - k)^2 = 3/4 - k.$

Sum: $3/4 - h + 3/4 - k = 3/2 - h - k = 0 \Rightarrow h + k = 3/2.$

Subtracting the second equation from the third:

$\left(\frac{x+1}{2} - h\right)^2 + \left(\frac{1-x}{2} - k\right)^2 - \left(\frac12 - h\right)^2 - \left(\frac12 - k\right)^2 = 0.$

Compute each: $\left(\frac{x+1}{2} - h\right)^2 - \left(\frac12 - h\right)^2 = \left(\frac{x}{2}\right)^2 + x\left(\frac12 - h\right) = \frac{x^2}{4} + \frac{x}{2} - x h = \frac{x^2}{4} + \frac{x}{2} - x h$. Similarly, $\left(\frac{1-x}{2} - k\right)^2 - \left(\frac12 - k\right)^2 = \left(-\frac{x}{2}\right)^2 - x \cdot \frac12? \dots$ Wait, carefully: $(\frac{1-x}{2} - k)^2 - (\frac12 - k)^2 = (\frac12 - k - \frac{x}{2})^2 - (\frac12 - k)^2 = (\frac12 - k)^2 - (\frac12 - k) x + x^2 /4 - (\frac12 - k)^2 = - (\frac12 - k) x + x^2 /4 = x^2 /4 - x(\frac12 - k)$. Sum of two differences: $(x^2/4 + x/2 - x h) + (x^2/4 - x/2 + x k) = x^2/2 + (-x h + x k) + 0 = x^2/2 + x(k - h) = 0 \Rightarrow k - h = -x/2 \Rightarrow k = h - x/2.$

Combine with $h + k = 3/2 \Rightarrow h + h - x/2 = 3/2 \Rightarrow 2h - x/2 = 3/2 \Rightarrow 2h = 3/2 + x/2 \Rightarrow h = \frac{3+x}{4} \Rightarrow k = h - x/2 = \frac{3 + x}{4} - \frac{x}{2} = \frac{3 - x}{4}.$

Thus, the circle center is $(h,k)=(\frac{3+x}{4}, \frac{3-x