Kvant Math Problem 1096
Let the circle have radius $R=\dfrac d2$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m21s
Source on kvant.digital
Problem
The diameter $d$ of a circle is divided into $k$ equal parts, and through each division point a chord perpendicular to the diameter is drawn. Prove that the sum of the lengths of all the chords is not less than $0{,}5kd$ and less than $0{,}8kd$.
R. Kharitonov, A. Chagirov, 10th-grade students
Exploration
Let the circle have radius $R=\dfrac d2$. The diameter is divided into $k$ equal parts, hence there are $k-1$ interior division points. Through each such point a chord perpendicular to the diameter is drawn. If the coordinate of the point is $x$, measured from the center along the diameter, then the chord length equals
$$2\sqrt{R^2-x^2}.$$
The division points are
$$x_i=-R+\frac{i}{k}d =R!\left(\frac{2i}{k}-1\right), \qquad i=1,\dots,k-1.$$
Hence the sum of all chord lengths is
$$S =2R\sum_{i=1}^{k-1} \sqrt{1-\left(\frac{2i}{k}-1\right)^2}.$$
Since $d=2R$,
$$S=d\sum_{i=1}^{k-1} \sqrt{1-\left(\frac{2i}{k}-1\right)^2}.$$
The expression under the sum is the value of the function
$$f(x)=\sqrt{1-x^2}$$
at equally spaced points of $[-1,1]$. The bounds $0.5kd$ and $0.8kd$ suggest comparison with the area under the semicircle,
$$\int_{-1}^{1}\sqrt{1-x^2},dx=\frac{\pi}{2}\approx1.57.$$
Indeed,
$$S=d\sum f(x_i),$$
while
$$\frac{2}{k}\sum f(x_i)$$
is a Riemann sum for that integral. Thus one expects
$$S\approx \frac{\pi}{4}kd\approx0.785,kd,$$
which explains the constant $0.8$.
To obtain rigorous inequalities, the shape of $f$ is crucial. Since
$$f''(x)=-(1-x^2)^{-3/2}<0,$$
$f$ is concave on $(-1,1)$. For a concave function, the graph lies above every chord and below every tangent. Consequently, on each subinterval the trapezoidal rule underestimates the integral, whereas the midpoint rule overestimates it. The lower bound should come from the trapezoidal estimate, the upper bound from the midpoint estimate.
The step most likely to hide an error is the conversion of these geometric facts about concave functions into exact inequalities for the particular sums occurring here.
Problem Understanding
The diameter of a circle of diameter $d$ is divided into $k$ equal segments. Through each interior division point a chord perpendicular to the diameter is drawn. If $S$ denotes the sum of the lengths of all these chords, prove
$$0.5,kd\le S<0.8,kd.$$
This is a Type B problem.
The core difficulty is to convert the discrete sum of chord lengths into a quantity that can be compared with the area of a semicircle and then obtain explicit numerical bounds. The relevant function is $f(x)=\sqrt{1-x^2}$, whose concavity permits comparison between its integral and suitable Riemann sums.
Proof Architecture
Let $f(x)=\sqrt{1-x^2}$ on $[-1,1]$.
Lemma 1. The sum of chord lengths satisfies
$$S=d\sum_{i=1}^{k-1}f!\left(-1+\frac{2i}{k}\right).$$
This follows directly from the formula for a chord at distance $|x|$ from the center.
Lemma 2. For every concave function on an interval, the trapezoidal rule underestimates the integral.
The graph of a concave function lies above the secant line joining the endpoints.
Lemma 3. Applied to $f$ on the partition of $[-1,1]$ into $k$ equal parts,
$$\frac{2}{k}\sum_{i=1}^{k-1}f!\left(-1+\frac{2i}{k}\right) \ge \int_{-1}^{1}f(x),dx .$$
The endpoint values vanish, so the trapezoidal estimate simplifies to the displayed sum.
Lemma 4. For every concave function, the midpoint rule overestimates the integral.
The graph lies below the tangent at the midpoint of each subinterval.
Lemma 5. Applied to the partition into $k-1$ equal parts whose nodes are exactly the sampling points of Lemma 1,
$$\frac{2}{k-1}\sum_{i=1}^{k-1}f!\left(-1+\frac{2i}{k}\right) < \int_{-1}^{1}f(x),dx .$$
Since $f$ is not linear, the inequality is strict.
The hardest direction is the upper bound $S<0.8kd$, which requires a strict estimate obtained from the midpoint rule.
Solution
Let the circle have radius
$$R=\frac d2.$$
Choose coordinates so that the diameter lies on the $x$-axis and the center of the circle is the origin. The circle is then
$$x^2+y^2=R^2.$$
The interior division points of the diameter have coordinates
$$x_i=-R+\frac{i}{k}d =R!\left(\frac{2i}{k}-1\right), \qquad i=1,\dots,k-1.$$
The chord through $x_i$ perpendicular to the diameter has endpoints on the circle, hence its length is
$$2\sqrt{R^2-x_i^2}.$$
Therefore
$$S = 2\sum_{i=1}^{k-1}\sqrt{R^2-x_i^2} = d\sum_{i=1}^{k-1} \sqrt{1-\left(\frac{2i}{k}-1\right)^2}.$$
Define
$$f(x)=\sqrt{1-x^2}, \qquad -1\le x\le1.$$
Then
$$S=d\sum_{i=1}^{k-1}f!\left(-1+\frac{2i}{k}\right).$$
Since
$$f''(x)=-(1-x^2)^{-3/2}<0 \qquad (-1<x<1),$$
$f$ is concave on $[-1,1]$.
Consider the partition
$$-1=x_0<x_1<\cdots<x_k=1, \qquad x_i=-1+\frac{2i}{k}.$$
For a concave function, on every interval $[x_{i-1},x_i]$ the graph lies above the secant joining the endpoints. Hence
$$\int_{x_{i-1}}^{x_i}f(x),dx \ge \frac{x_i-x_{i-1}}2 \bigl(f(x_{i-1})+f(x_i)\bigr).$$
Summing over all $i$ and using $x_i-x_{i-1}=\dfrac2k$ gives
$$\int_{-1}^{1}f(x),dx \ge \frac1k \left( f(-1)+2\sum_{i=1}^{k-1}f(x_i)+f(1) \right).$$
Since $f(-1)=f(1)=0$,
$$\int_{-1}^{1}f(x),dx \ge \frac2k\sum_{i=1}^{k-1}f(x_i).$$
The integral equals the area of the upper semicircle of radius $1$:
$$\int_{-1}^{1}f(x),dx=\frac{\pi}{2}.$$
Thus
$$\sum_{i=1}^{k-1}f(x_i) \le \frac{\pi k}{4},$$
and consequently
$$S\le \frac{\pi}{4}kd.$$
Since
$$\frac{\pi}{4}<0.8,$$
we obtain
$$S<0.8,kd.$$
For the lower bound, divide $[-1,1]$ into $k-1$ equal intervals with endpoints
$$y_j=-1+\frac{2j}{k-1}, \qquad j=0,\dots,k-1.$$
Their midpoints are
$$m_j=-1+\frac{2j+1}{k-1}.$$
Because $f$ is concave, on each interval the graph lies below the tangent at the midpoint, hence
$$\int_{y_j}^{y_{j+1}}f(x),dx \le \frac{2}{k-1},f(m_j).$$
The inequality is strict because $f$ is not linear on any interval. Summing yields
$$\frac{\pi}{2} < \frac{2}{k-1}\sum_{j=0}^{k-2}f(m_j).$$
The points $m_j$ are exactly
$$-1+\frac{2i}{k}, \qquad i=1,\dots,k-1,$$
so
$$\frac{\pi}{2} < \frac{2}{k-1} \sum_{i=1}^{k-1}f(x_i).$$
Therefore
$$\sum_{i=1}^{k-1}f(x_i) > \frac{\pi}{4}(k-1).$$
Multiplying by $d$,
$$S>\frac{\pi}{4}(k-1)d.$$
Since $\dfrac{\pi}{4}>0.78$ and $k-1\ge \dfrac{2}{3}k$ for every $k\ge2$,
$$S>\frac{\pi}{4}(k-1)d > 0.52,kd > 0.5,kd.$$
Hence
$$0.5,kd<S<0.8,kd.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the formula for a chord. At distance $|x|$ from the center, the circle equation gives
$$y=\pm\sqrt{R^2-x^2}.$$
The vertical distance between the two intersection points is
$$2\sqrt{R^2-x^2}.$$
No approximation is involved.
The second delicate step is the trapezoidal estimate. For a concave function, the secant joining $(a,f(a))$ and $(b,f(b))$ lies below the graph. Integrating over $[a,b]$ gives
$$\int_a^b f(x),dx \ge \frac{b-a}{2}\bigl(f(a)+f(b)\bigr).$$
Reversing this inequality would produce the wrong direction for the upper bound.
The third delicate step is the midpoint estimate. For a concave function, the tangent at the midpoint lies above the graph. The integral of that tangent over a symmetric interval about the midpoint equals the interval length multiplied by the midpoint value. Thus
$$\int_a^b f(x),dx \le (b-a)f!\left(\frac{a+b}{2}\right).$$
Because $f(x)=\sqrt{1-x^2}$ is strictly concave, equality cannot occur on a nondegenerate interval, which yields the strict upper bound $S<0.8kd$.
Alternative Approaches
A geometric proof can be formulated without explicitly introducing integrals. The chords correspond to ordinates of the upper semicircle. The sum of their lengths is twice the sum of these ordinates. One inscribes and circumscribes the semicircle by polygons obtained from the same equally spaced vertical lines. The areas of the resulting trapezoids and rectangles provide lower and upper estimates for the area of the semicircle. Translating those area inequalities gives the same bounds for the chord sum.
The analytic approach with the concave function $f(x)=\sqrt{1-x^2}$ is preferable because both inequalities emerge from standard integral estimates, and the constants arise directly from the semicircle area $\pi/2$.