Kvant Math Problem 788
Let
Verified: yes
Verdicts: PASS + PASS
Solve time: 9m34s
Source on kvant.digital
Problem
- On the graph $y=x^2$, the points $A(a;,a^2)$ and $B(b;,b^2)$ are marked. Find a point $M(m;,m^2)$ between them for which the sum of the areas of the two segments bounded by the graph and the chords $AM$ and $BM$ is minimal.
- On the graph of a differentiable function $y=f(x)$, the points $A$ and $B$ are marked. It is known that the graph and the chord $AB$ bound a convex figure. Let $M$ be a point on the graph lying between $A$ and $B$ for which the sum of the areas of the two segments bounded by the graph and the chords $AM$ and $BM$ is minimal. Prove that the tangent to the graph at the point $M$ is parallel to the chord $AB$.
E. D. Moskalensky, Yu. V. Chinyaev
Exploration
Let
$$A=(a,a^2),\qquad B=(b,b^2),\qquad a<b,$$
and let
$$M=(m,m^2),\qquad a<m<b.$$
For the parabola, the area of the segment cut off by a chord can be computed explicitly. The chord through points with abscissas $u<v$ has equation
$$y=(u+v)x-uv.$$
Hence
$$x^2-\bigl((u+v)x-uv\bigr) =(x-u)(x-v).$$
On $[u,v]$ this quantity is nonpositive, so the area between the chord and the parabola equals
$$\int_u^v \bigl((u+v)x-uv-x^2\bigr),dx =\int_u^v (x-u)(v-x),dx.$$
Setting $L=v-u$ and writing $x=u+t$,
$$\int_0^L t(L-t),dt =\frac{L^3}{6}.$$
Thus the segment determined by a chord joining two points of the parabola depends only on the horizontal distance of the endpoints.
For the present problem the quantity to minimize is
$$S(m)=\frac{(m-a)^3}{6}+\frac{(b-m)^3}{6}.$$
Differentiating gives
$$S'(m)=\frac12\bigl((m-a)^2-(b-m)^2\bigr).$$
The critical point satisfies
$$m-a=b-m,$$
hence
$$m=\frac{a+b}{2}.$$
Since
$$S''(m)=(m-a)+(b-m)=b-a>0,$$
this point indeed gives the minimum.
For the second part it is natural to express the total area as the area between the graph and the broken line $A!-!M!-!B$. The area between the graph and the fixed chord $AB$ is constant, so minimizing the sum of the two segment areas is equivalent to maximizing the area enclosed by the broken line $A!-!M!-!B$ and the chord $AB$, namely the area of triangle $AMB$.
The base $AB$ is fixed. Therefore we must maximize the distance from $M$ to the line $AB$. Since the graph and the chord bound a convex figure, every point of the graph between $A$ and $B$ lies on the same side of $AB$, and maximizing the distance is equivalent to maximizing the vertical difference between the graph and the line $AB$.
If
$$\ell(x)$$
is the equation of the chord $AB$, then we maximize
$$g(x)=f(x)-\ell(x).$$
Because $g(a)=g(b)=0$ and $g\ge0$, the maximum occurs at an interior point $m$. Differentiability gives
$$g'(m)=0,$$
hence
$$f'(m)=\ell'(x).$$
The slope of $\ell$ is exactly the slope of $AB$, so the tangent at $M$ is parallel to $AB$.
The delicate point is the equivalence between minimizing the sum of the two segment areas and maximizing the area of triangle $AMB$.
Problem Understanding
This is a Type C problem.
In the first part we must determine the point $M$ on the parabola $y=x^2$ lying between two given points $A$ and $B$ for which the sum of the areas of the two parabolic segments cut off by chords $AM$ and $BM$ is as small as possible.
In the second part the parabola is replaced by an arbitrary differentiable curve whose graph and the chord $AB$ bound a convex figure. We must prove that any point $M$ minimizing the analogous sum of segment areas is characterized by the fact that the tangent at $M$ is parallel to $AB$.
The core difficulty is to relate the sum of the two segment areas to a simpler geometric quantity depending on $M$.
For the first part the answer is
$$m=\frac{a+b}{2}.$$
The reason is that the area of a parabolic segment determined by two points of $y=x^2$ is proportional to the cube of the horizontal distance between the endpoints, reducing the problem to minimizing a simple cubic expression.
Proof Architecture
First, compute the area of the segment bounded by the parabola $y=x^2$ and a chord joining two points with abscissas $u$ and $v$; it equals $\frac{(v-u)^3}{6}$.
Next, express the total area in part 1 as
$$S(m)=\frac{(m-a)^3}{6}+\frac{(b-m)^3}{6}.$$
Then show that $S'(m)=0$ exactly when $m=\frac{a+b}{2}$, and that $S''(m)>0$, yielding the minimum.
For part 2, prove that the sum of the two segment areas equals a fixed area minus the area of triangle $AMB$.
Then show that minimizing the sum is equivalent to maximizing the area of triangle $AMB$.
Express the area of triangle $AMB$ as one half of the fixed length $|AB|$ times the distance from $M$ to the line $AB$, so maximizing the triangle area is equivalent to maximizing the distance from $M$ to $AB$.
Finally, if $\ell(x)$ is the equation of the chord $AB$, show that maximizing the distance is equivalent to maximizing $f(x)-\ell(x)$; differentiability then gives $f'(m)=\ell'$, which means that the tangent at $M$ is parallel to $AB$.
The most delicate lemma is the identity expressing the sum of the two segment areas as a fixed quantity minus the area of triangle $AMB$.
Solution
Assume throughout that $a<b$.
For two points
$$U=(u,u^2),\qquad V=(v,v^2),$$
with $u<v$, the chord $UV$ has slope
$$\frac{v^2-u^2}{v-u}=u+v,$$
hence its equation is
$$y=(u+v)x-uv.$$
The area $T(u,v)$ of the segment bounded by this chord and the parabola is
$$T(u,v) =\int_u^v\bigl((u+v)x-uv-x^2\bigr),dx.$$
Since
$$(u+v)x-uv-x^2 =(x-u)(v-x),$$
we obtain
$$T(u,v) =\int_u^v (x-u)(v-x),dx.$$
Put $x=u+t$, $L=v-u$. Then
$$T(u,v) =\int_0^L t(L-t),dt =L\int_0^L t,dt-\int_0^L t^2,dt =\frac{L^3}{2}-\frac{L^3}{3} =\frac{L^3}{6}.$$
Therefore
$$T(u,v)=\frac{(v-u)^3}{6}.$$
For
$$M=(m,m^2),$$
the required sum of areas is
$$S(m) =T(a,m)+T(m,b) =\frac{(m-a)^3}{6}+\frac{(b-m)^3}{6}.$$
Differentiating,
$$S'(m) =\frac12\Bigl((m-a)^2-(b-m)^2\Bigr).$$
The equation $S'(m)=0$ is
$$(m-a)^2=(b-m)^2.$$
Since $a<m<b$,
$$m-a=b-m,$$
and therefore
$$m=\frac{a+b}{2}.$$
Moreover,
$$S''(m) =(m-a)+(b-m)=b-a>0.$$
Hence $S$ is strictly convex and the critical point is the unique minimum.
Thus in part 1 the minimizing point is
$$\boxed{M!\left(\frac{a+b}{2},,\left(\frac{a+b}{2}\right)^2\right)}.$$
Now consider part 2.
Let $R$ be the convex figure bounded by the graph of $y=f(x)$ and the chord $AB$. Its area is fixed.
Let $S(M)$ denote the sum of the areas of the two segments bounded respectively by the graph and the chords $AM$ and $MB$.
The union of these two segments together with the triangle $AMB$ is exactly the figure $R$. Their interiors do not overlap. Consequently,
$$\operatorname{Area}(R) = S(M)+\operatorname{Area}(\triangle AMB).$$
Since $\operatorname{Area}(R)$ is independent of $M$, minimizing $S(M)$ is equivalent to maximizing
$$\operatorname{Area}(\triangle AMB).$$
The side $AB$ is fixed. Hence
$$\operatorname{Area}(\triangle AMB) = \frac12,|AB|,d(M,AB),$$
where $d(M,AB)$ is the distance from $M$ to the line through $AB$. Therefore $S(M)$ is minimal precisely when $d(M,AB)$ is maximal.
Let
$$y=\ell(x)$$
be the equation of the chord $AB$. Because the graph and the chord bound a convex figure, every point of the graph between $A$ and $B$ lies on the same side of the line $AB$. Thus maximizing the distance from $M=(x,f(x))$ to $AB$ is equivalent to maximizing
$$g(x)=f(x)-\ell(x).$$
Since $g(a)=g(b)=0$ and $g\ge0$ on $[a,b]$, a maximum is attained at some interior point $m$. Because $f$ is differentiable and $\ell$ is linear, $g$ is differentiable. At an interior maximum,
$$g'(m)=0.$$
Hence
$$f'(m)-\ell'(m)=0,$$
so
$$f'(m)=\ell'.$$
The number $\ell'$ is the slope of the chord $AB$. Thus the tangent to the graph at
$$M=(m,f(m))$$
has the same slope as $AB$, which means that the tangent at $M$ is parallel to the chord $AB$.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the computation of the parabolic segment area. Starting from
$$y=(u+v)x-uv,$$
the difference between chord and parabola is
$$(u+v)x-uv-x^2=(x-u)(v-x).$$
Integrating directly,
$$\int_u^v (x-u)(v-x),dx =\int_0^{v-u} t((v-u)-t),dt =\frac{(v-u)^3}{6}.$$
No sign error is possible because the integrand is nonnegative on the interval.
The second delicate step is
$$\operatorname{Area}(R) = S(M)+\operatorname{Area}(\triangle AMB).$$
The figure $R$ is partitioned by the chords $AM$ and $MB$ into exactly three regions: the segment adjacent to $AM$, the segment adjacent to $MB$, and the triangle $AMB$. Any omission or overlap would contradict the fact that the boundaries of these regions consist precisely of pieces of the graph and the two chords.
The third delicate step is the passage from maximal distance to the derivative condition. If the line $AB$ has equation $\ell(x)$, then distance from $(x,f(x))$ to this line equals
$$\frac{|f(x)-\ell(x)|}{\sqrt{1+\ell'^2}}.$$
All points of the graph lie on the same side of the line, so the absolute value may be removed. Since the denominator is constant, maximizing distance is exactly the same as maximizing $f(x)-\ell(x)$. The interior maximum condition then yields
$$f'(m)=\ell'.$$
Alternative Approaches
For part 1 one may avoid differentiation. Writing
$$x=m-a,\qquad y=b-m,$$
we have $x+y=b-a$ and
$$S(m)=\frac{x^3+y^3}{6}.$$
Using
$$x^3+y^3=(x+y)^3-3xy(x+y),$$
the sum is minimized when $xy$ is maximal. For fixed $x+y$, the product $xy$ is maximal at $x=y$, yielding $m=\frac{a+b}{2}$.
For part 2 one may introduce coordinates with the chord $AB$ as the $x$-axis. Then the area of triangle $AMB$ is proportional to the ordinate of $M$. The minimizing point is therefore the highest point of the graph in the new coordinate system. At such a point the tangent is horizontal, which means, in the original coordinates, that the tangent is parallel to $AB$. The main solution is preferable because it keeps the geometry transparent and works directly with the original coordinates.