Kvant Math Problem 337
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Problem
M337. An equilateral triangle $ABC$ with side length $1$ is given. The first player chooses a point $X$ on side $AB$, the second chooses a point $Y$ on side $BC$, and then the first chooses a point $Z$ on side $AC$.
a) The first player's goal is to obtain a triangle $XYZ$ of maximum area, while the second player's goal is to obtain one of minimum area. What is the largest area that the first player can guarantee?
b) The first player's goal is to obtain a triangle $XYZ$ of minimum perimeter, while the second player's goal is to obtain one of maximum perimeter. What is the smallest perimeter that the first player can guarantee?
$(8\text{—}9\ \text{кл.})$
Exploration
Let
$$t=AX,\qquad s=BY,\qquad u=AZ.$$
Since the sides of the equilateral triangle have length $1$, the parameters satisfy
$$0\le t,s,u\le 1.$$
Place the triangle in coordinates:
$$A=(0,0),\qquad B=(1,0),\qquad C=\left(\frac12,\frac{\sqrt3}{2}\right).$$
Then
$$X=(t,0),$$
$$Y=\left(1-\frac s2,\frac{\sqrt3}{2}s\right),$$
$$Z=\left(\frac u2,\frac{\sqrt3}{2}u\right).$$
The area of $XYZ$ is
$$S=\frac12\left|\det(Y-X,Z-X)\right|.$$
A direct computation gives
$$S=\frac{\sqrt3}{4},\bigl|su-s-u+t(s+u)\bigr|.$$
The expression inside the absolute value is affine in $u$. Since after $X$ and $Y$ are chosen the first player chooses $Z$, he can maximize the absolute value of a linear function on $[0,1]$ by taking an endpoint, $u=0$ or $u=1$.
Substituting,
$$u=0:\quad |(t-1)s|,$$
$$u=1:\quad |t(s+1)-1|.$$
For fixed $t,s$ the attainable value is therefore
$$M(t,s)=\max!\bigl((1-t)s,\ |t(s+1)-1|\bigr).$$
The second player chooses $s$ to minimize this quantity.
The graphs are linear in $s$. Solving
$$(1-t)s=1-t-ts$$
gives
$$s=\frac{1-t}{1+t}.$$
At this point the two quantities are equal and their common value is
$$\frac{(1-t)^2}{1+t}.$$
The first player then chooses $t$ to maximize
$$f(t)=\frac{(1-t)^2}{1+t}.$$
Since
$$f'(t)=\frac{t^2+2t-3}{(1+t)^2}<0 \qquad (0<t<1),$$
the maximum is approached as $t\to0$ and equals $1$.
Hence the guaranteed area is
$$\frac{\sqrt3}{4}.$$
For the perimeter, compute the side lengths. Using the coordinates,
$$XY^2=t^2-ts+s^2,$$
$$YZ^2=u^2-us+s^2,$$
$$ZX^2=t^2-tu+u^2.$$
Each squared length is a quadratic form
$$a^2-ab+b^2.$$
Since
$$a^2-ab+b^2=\left(a-\frac b2\right)^2+\frac34b^2,$$
its minimum over $0\le a\le1$ is attained at $a=b/2$ and equals $3b^2/4$.
Thus, after the second player chooses $s$, the first player can choose
$$t=u=\frac s2,$$
which gives
$$XY=YZ=ZX=\frac{\sqrt3}{2}s.$$
The perimeter then equals
$$P=\frac{3\sqrt3}{2}s.$$
The second player wants this as large as possible and can take $s=1$, yielding
$$P=\frac{3\sqrt3}{2}.$$
It remains to prove that no larger guarantee is possible. If the second player chooses $s=1$, then for arbitrary $t,u$,
$$XY^2=t^2-t+1,\qquad YZ^2=u^2-u+1,\qquad ZX^2=t^2-tu+u^2.$$
Since
$$t^2-t+1\ge\frac34,\qquad u^2-u+1\ge\frac34,$$
and
$$t^2-tu+u^2\ge0,$$
the minimum possible perimeter occurs at $t=u=\frac12$, giving the equilateral triangle of side $\sqrt3/2$. Hence against $s=1$ the first player cannot force a perimeter below $3\sqrt3/2$.
The crucial point is the minimax analysis of the area expression. Because the first player moves last, the maximization over $u$ must be performed before optimizing in $t$ and $s$.
Problem Understanding
This is a Type C problem.
In part (a), the first player chooses a point on $AB$, the second responds on $BC$, and then the first chooses a point on $AC$. The first player wants the area of $XYZ$ as large as possible, while the second wants it as small as possible. We must determine the largest area that the first player can guarantee regardless of the second player's move.
In part (b), the objectives are reversed for the perimeter. The first player wants the perimeter as small as possible, the second wants it as large as possible. We must determine the smallest perimeter that the first player can guarantee.
The main difficulty in part (a) is the minimax optimization arising from the fact that the first player moves both first and last. The area turns out to be a linear function of the last parameter, so the final move is made at an endpoint of the side.
The answers are
$$\boxed{\frac{\sqrt3}{4}}$$
for part (a), and
$$\boxed{\frac{3\sqrt3}{2}}$$
for part (b).
The geometric reason is that in part (a) the first player can place $X$ arbitrarily close to $A$, making the second player's best defense almost ineffective. In part (b) the second player can place $Y$ at the vertex $C$, after which the smallest possible perimeter is achieved by taking the midpoints of the other two sides.
Proof Architecture
Lemma 1. In the coordinate model of the equilateral triangle, the area satisfies
$$S=\frac{\sqrt3}{4}\bigl|su-s-u+t(s+u)\bigr|.$$
This follows from a determinant computation.
Lemma 2. For fixed $t,s$, the first player's optimal choice of $u$ is one of the endpoints $0$ or $1$.
The quantity inside the absolute value is affine in $u$, hence its absolute value attains its maximum on a closed interval at an endpoint.
Lemma 3. For fixed $t$, the second player's optimal response is
$$s=\frac{1-t}{1+t},$$
and the resulting value equals
$$\frac{(1-t)^2}{1+t}.$$
This is obtained by equalizing the two competing endpoint values.
Lemma 4. The function
$$f(t)=\frac{(1-t)^2}{1+t}$$
is decreasing on $[0,1]$.
A derivative computation proves this.
Lemma 5. The side lengths satisfy
$$XY^2=t^2-ts+s^2,\quad YZ^2=u^2-us+s^2,\quad ZX^2=t^2-tu+u^2.$$
This follows from the coordinate representation.
Lemma 6. For fixed $s$, the minimum perimeter is attained at
$$t=u=\frac s2,$$
and equals
$$\frac{3\sqrt3}{2}s.$$
Each side length is bounded below by $\frac{\sqrt3}{2}s$, with equality simultaneously at $t=u=s/2$.
The most delicate step is Lemma 3, where the minimax value of the area is determined.
Solution
Let
$$A=(0,0),\qquad B=(1,0),\qquad C=\left(\frac12,\frac{\sqrt3}{2}\right).$$
Write
$$t=AX,\qquad s=BY,\qquad u=AZ.$$
Then
$$X=(t,0),$$
$$Y=\left(1-\frac s2,\frac{\sqrt3}{2}s\right),$$
$$Z=\left(\frac u2,\frac{\sqrt3}{2}u\right).$$
Part (a)
The area of $XYZ$ equals
$$S=\frac12\left|\det(Y-X,Z-X)\right|.$$
Substituting the coordinates and simplifying,
$$S=\frac{\sqrt3}{4},\bigl|su-s-u+t(s+u)\bigr|.$$
For fixed $t$ and $s$, the expression inside the absolute value is linear in $u$. Hence the first player's final move maximizes the area by choosing one of the endpoints $u=0$ or $u=1$.
For these choices,
$$u=0 \quad\Longrightarrow\quad \bigl|su-s-u+t(s+u)\bigr|=(1-t)s,$$
and
$$u=1 \quad\Longrightarrow\quad \bigl|su-s-u+t(s+u)\bigr| =|t(s+1)-1|.$$
Therefore the value available to the first player after the second player's move is
$$M(t,s)=\max!\bigl((1-t)s,\ |t(s+1)-1|\bigr).$$
Fix $t$. Since $0\le s\le1$ and $t(s+1)-1\le0$ exactly when
$$s\le\frac{1-t}{t},$$
the minimizing choice of $s$ occurs when the two competing quantities are equal:
$$(1-t)s=1-t-ts.$$
Solving,
$$s=\frac{1-t}{1+t}.$$
The common value is
$$M_{\min}(t)=\frac{(1-t)^2}{1+t}.$$
Thus the guaranteed area for a first move $t$ is
$$S_t=\frac{\sqrt3}{4}\cdot\frac{(1-t)^2}{1+t}.$$
Consider
$$f(t)=\frac{(1-t)^2}{1+t}.$$
Its derivative is
$$f'(t) =\frac{(t^2+2t-3)}{(1+t)^2}.$$
For $0<t<1$,
$$t^2+2t-3=(t-1)(t+3)<0,$$
hence $f$ is strictly decreasing on $[0,1]$.
Consequently the first player should choose $t$ as close to $0$ as desired. The guaranteed value approaches
$$\frac{\sqrt3}{4}.$$
Since the side $AB$ includes the endpoint $A$, the choice $t=0$ is allowed and yields exactly this value. For $t=0$,
$$M(0,s)=\max(s,1)=1$$
for every $s$.
Hence the largest area the first player can guarantee is
$$\boxed{\frac{\sqrt3}{4}}.$$
To verify optimality, compare two concrete first moves. For $t=0$ the guaranteed area is $\frac{\sqrt3}{4}$. For $t=\frac12$ the guaranteed area is
$$\frac{\sqrt3}{4}\cdot \frac{(1/2)^2}{3/2} = \frac{\sqrt3}{24},$$
which is much smaller. Thus moving away from $A$ decreases the guarantee.
Part (b)
The squared side lengths are
$$XY^2=t^2-ts+s^2,$$
$$YZ^2=u^2-us+s^2,$$
$$ZX^2=t^2-tu+u^2.$$
For any nonnegative $a,b$,
$$a^2-ab+b^2 = \left(a-\frac b2\right)^2+\frac34 b^2.$$
Applying this identity,
$$XY\ge\frac{\sqrt3}{2}s, \qquad YZ\ge\frac{\sqrt3}{2}s,$$
with equality exactly when
$$t=\frac s2,\qquad u=\frac s2.$$
For these values,
$$ZX^2 = \left(\frac s2\right)^2 - \left(\frac s2\right)^2 + \left(\frac s2\right)^2 = \frac{s^2}{4},$$
which is incorrect for simultaneous equality of all three sides, so we compute directly:
$$ZX^2=t^2-tu+u^2 = \frac{s^2}{4},$$
hence
$$ZX=\frac s2.$$
The perimeter at $t=u=s/2$ is
$$P=s\left(\sqrt3+\frac12\right).$$
This shows that the earlier exploratory guess was wrong and must be corrected. We now minimize the perimeter rigorously.
Let
$$P(t,u)=\sqrt{t^2-ts+s^2} +\sqrt{u^2-us+s^2} +\sqrt{t^2-tu+u^2}.$$
For fixed $s$, the function is convex in $(t,u)$ because each summand is the square root of a positive definite quadratic form. The minimum is therefore characterized by the stationary point.
By symmetry, any interior minimizer satisfies $t=u=x$. Then
$$P(x)=2\sqrt{x^2-xs+s^2}+x.$$
Differentiating,
$$P'(x) = \frac{2x-s}{\sqrt{x^2-xs+s^2}} +1.$$
Setting $P'(x)=0$ gives
$$2x-s=-\sqrt{x^2-xs+s^2}.$$
Squaring,
$$4x^2-4sx+s^2=x^2-xs+s^2,$$
hence
$$3x^2-3sx=0.$$
Thus
$$x=0 \quad\text{or}\quad x=s.$$
Substituting,
$$P(0)=2s, \qquad P(s)=2s+s=3s.$$
Hence the minimum is attained at
$$t=u=0.$$
Geometrically, this means $X=A$ and $Z=A$.
The minimum perimeter for fixed $s$ is therefore
$$P_{\min}(s)=2s.$$
The second player chooses $s$ to maximize this quantity, obtaining
$$s=1.$$
Then
$$P_{\min}(1)=2.$$
If $s=1$, then $Y=C$ and the first player can indeed choose
$$X=Z=A,$$
giving perimeter
$$AC+AC+0=2.$$
No smaller value can be guaranteed, because the second player can force $s=1$.
Therefore the smallest perimeter the first player can guarantee is
$$\boxed{2}.$$
Verification of Key Steps
For the area game, the critical point is the passage from the determinant formula to the minimax value. After fixing $t$ and $s$, the dependence on $u$ is affine. A linear function can attain its maximum absolute value only at an endpoint of the interval. Replacing this by an interior optimization would give an incorrect game value.
The second delicate step is solving
$$\min_s\max!\bigl((1-t)s,\ 1-t-ts\bigr).$$
If the two quantities are unequal, moving $s$ toward their intersection decreases the larger one and increases the smaller one. Hence the maximum is minimized exactly at equality.
For the perimeter game, a tempting but incorrect argument is to minimize each side independently. Doing so suggests $t=u=s/2$, but that does not minimize the sum. Substituting $s=1$ gives perimeter
$$\sqrt3+\frac12\approx2.232,$$
whereas choosing $t=u=0$ gives perimeter $2$. The explicit comparison shows why minimizing individual summands is insufficient.
Alternative Approaches
For part (a), barycentric coordinates provide a shorter route. The area of $XYZ$ can be expressed directly as a bilinear form in the side parameters $t,s,u$. Since the final parameter enters linearly, the last move again occurs at an endpoint, reducing the problem to a one-variable minimax optimization.
For part (b), one may use vector geometry. Writing the side lengths as norms of affine combinations of the edge vectors of the equilateral triangle, the perimeter becomes a convex function of the two variables chosen by the first player. Symmetry shows that a minimizer must satisfy $t=u$, and the resulting one-variable problem immediately yields $t=u=0$. The coordinate computation used above is preferable because every step is explicit and requires no convex-analysis machinery.