Kvant Math Problem 263
Let the rectangle have coordinates
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m48s
Source on kvant.digital
Problem
Two numbers $p$ and $q$ greater than 1 are given. On the sides $BC$ and $DC$ of the rectangle $ABCD$, points $P$ and $Q$ are chosen so that $|BC| = p \cdot |BP|$ and $|DC| = q \cdot |DQ|$. For what ratio of the lengths of the sides $AB$ and $AD$ will the angle $PAQ$ attain its greatest value? What is this greatest value in the particular case $p = 2$, $q = \dfrac{3}{2}$ (Fig. 2)?
Figure 2
E. G. Gotman
Exploration
Let the rectangle have coordinates
$$A=(0,0),\quad B=(a,0),\quad D=(0,b),\quad C=(a,b),$$
where $a=|AB|$ and $b=|AD|$.
Since
$$|BC|=b=p\cdot |BP|,$$
we have
$$|BP|=\frac b p,$$
hence
$$P=\left(a,\frac b p\right).$$
Similarly,
$$|DC|=a=q\cdot |DQ|,$$
so
$$|DQ|=\frac a q,$$
and therefore
$$Q=\left(\frac a q,b\right).$$
The angle to be maximized is the angle between the vectors
$$\overrightarrow{AP}=\left(a,\frac b p\right),\qquad \overrightarrow{AQ}=\left(\frac a q,b\right).$$
Introduce
$$r=\frac ab>0.$$
The problem reduces to maximizing $\angle PAQ$ as a function of $r$.
The cosine of the angle is
$$\cos\theta = \frac{\left(a,\frac b p\right)\cdot\left(\frac a q,b\right)} {\sqrt{a^2+\frac{b^2}{p^2}}, \sqrt{\frac{a^2}{q^2}+b^2}}.$$
After dividing numerator and denominator by $b^2$,
$$\cos\theta = \frac{\frac{r^2}{q}+\frac1p} {\sqrt{r^2+\frac1{p^2}}, \sqrt{\frac{r^2}{q^2}+1}}.$$
Since $0<\theta<\pi/2$, maximizing $\theta$ is equivalent to minimizing $\cos\theta$.
The expression suggests using the Cauchy inequality. Indeed,
$$\frac{r^2}{q}+\frac1p = \left(r,\frac1p\right)\cdot \left(\frac r q,1\right).$$
The denominator is the product of the lengths of these two vectors. Hence $\cos\theta$ is exactly the cosine of the angle between
$$u=\left(r,\frac1p\right),\qquad v=\left(\frac r q,1\right).$$
Thus minimizing $\cos\theta$ means maximizing the angle between $u$ and $v$.
The slopes of these vectors are
$$m_u=\frac1{pr},\qquad m_v=\frac q r.$$
Their angle equals
$$\arctan\frac q r-\arctan\frac1{pr}.$$
Now the problem becomes one-variable calculus. Differentiating,
$$\frac{d\theta}{dr} = -\frac q{r^2+q^2} + \frac p{p^2r^2+1}.$$
Setting this equal to $0$,
$$\frac p{p^2r^2+1} = \frac q{r^2+q^2}.$$
This yields
$$p(r^2+q^2)=q(p^2r^2+1),$$
hence
$$(p-p^2q)r^2=q-pq^2.$$
Factoring,
$$p(1-pq)r^2=q(1-pq).$$
Since $p,q>1$, $pq\neq1$, so
$$pr^2=q, \qquad r^2=\frac q p.$$
This is a natural candidate. Checking the derivative shows it changes from positive to negative there, so the angle is maximal.
Substituting $r=\sqrt{q/p}$,
$$\theta_{\max} = \arctan\sqrt{pq} - \arctan\frac1{\sqrt{pq}}.$$
Since the two arguments are reciprocal and positive,
$$\theta_{\max} = 2\arctan\sqrt{pq}-\frac\pi2.$$
For $p=2$, $q=\frac32$,
$$pq=3,$$
so
$$\theta_{\max} = 2\arctan\sqrt3-\frac\pi2 = 2\cdot\frac\pi3-\frac\pi2 = \frac\pi6.$$
Problem Understanding
The rectangle $ABCD$ has side lengths $a=|AB|$ and $b=|AD|$. Point $P$ lies on $BC$ so that $BP=\frac1p,BC$, and point $Q$ lies on $DC$ so that $DQ=\frac1q,DC$. The ratio $a:b$ may vary.
We must determine for which ratio $a:b$ the angle $\angle PAQ$ is as large as possible, and then compute that largest angle when $p=2$ and $q=\frac32$.
This is a Type C problem. The quantity to be optimized is the angle $\angle PAQ$.
The core difficulty is expressing the angle in a form whose dependence on the side ratio $a/b$ can be analyzed. The decisive observation is that the angle may be written as the difference of two arctangent functions.
The expected answer is
$$\frac{|AB|}{|AD|}=\sqrt{\frac q p},$$
because at this ratio the two terms arising in the derivative balance exactly. The maximal angle is
$$2\arctan\sqrt{pq}-\frac{\pi}{2}.$$
Proof Architecture
Let $a=|AB|$, $b=|AD|$, and $r=a/b$.
The first claim is that
$$P=\left(a,\frac b p\right),\qquad Q=\left(\frac a q,b\right)$$
in a suitable coordinate system; this follows directly from the defining conditions on $P$ and $Q$.
The second claim is that
$$\angle PAQ = \arctan\frac q r-\arctan\frac1{pr};$$
this follows from interpreting $\angle PAQ$ as the angle between two vectors and computing their directions.
The third claim is that
$$\frac{d}{dr}\angle PAQ = -\frac q{r^2+q^2} +\frac p{p^2r^2+1};$$
this is obtained by differentiating the expression from the second claim.
The fourth claim is that the unique critical point satisfies
$$r^2=\frac q p;$$
this follows from solving the equation obtained by setting the derivative equal to zero.
The fifth claim is that the derivative is positive for
$$r<\sqrt{\frac q p}$$
and negative for
$$r>\sqrt{\frac q p},$$
so the critical point yields the global maximum.
The hardest step is proving that the critical point indeed gives the global maximum. The lemma most likely to fail under scrutiny is the sign analysis of the derivative.
Solution
Place the rectangle in the coordinate plane so that
$$A=(0,0),\quad B=(a,0),\quad D=(0,b),\quad C=(a,b),$$
where
$$a=|AB|,\qquad b=|AD|.$$
Since
$$|BC|=b=p,|BP|,$$
we obtain
$$|BP|=\frac b p.$$
Because $P$ lies on $BC$,
$$P=\left(a,\frac b p\right).$$
Similarly,
$$|DC|=a=q,|DQ|,$$
hence
$$|DQ|=\frac a q,$$
and therefore
$$Q=\left(\frac a q,b\right).$$
Let
$$r=\frac ab.$$
The vectors
$$\overrightarrow{AP} = \left(a,\frac b p\right), \qquad \overrightarrow{AQ} = \left(\frac a q,b\right)$$
have slopes
$$\frac{b/p}{a}=\frac1{pr}, \qquad \frac{b}{a/q}=\frac q r.$$
Since $p,q>1$, the inequality
$$\frac q r>\frac1{pr}$$
holds for every $r>0$. Thus the angle between the two vectors equals the difference of their direction angles:
$$\theta=\angle PAQ = \arctan\frac q r-\arctan\frac1{pr}.$$
Differentiate with respect to $r$:
$$\theta'(r) = -\frac q{r^2+q^2} +\frac p{p^2r^2+1}.$$
The critical points satisfy
$$-\frac q{r^2+q^2} +\frac p{p^2r^2+1}=0,$$
or
$$p(r^2+q^2)=q(p^2r^2+1).$$
After rearranging,
$$p(1-pq)r^2=q(1-pq).$$
Because $p,q>1$, we have $pq>1$, so $1-pq\neq0$. Hence
$$pr^2=q,$$
that is,
$$r^2=\frac q p.$$
To determine the sign of the derivative, write
$$\theta'(r) = \frac{p(r^2+q^2)-q(p^2r^2+1)} {(r^2+q^2)(p^2r^2+1)}.$$
The denominator is positive. The numerator equals
$$p(1-pq)r^2+q(pq-1) = (pq-1)(q-pr^2).$$
Since $pq-1>0$,
$$\operatorname{sgn}\theta'(r) = \operatorname{sgn}(q-pr^2).$$
Therefore
$$\theta'(r)>0 \quad\text{for}\quad r<\sqrt{\frac q p},$$
and
$$\theta'(r)<0 \quad\text{for}\quad r>\sqrt{\frac q p}.$$
The function $\theta(r)$ increases up to
$$r=\sqrt{\frac q p}$$
and decreases afterwards. Consequently the maximum is attained precisely when
$$\frac{|AB|}{|AD|} = \sqrt{\frac q p}.$$
At this value,
$$\frac q r=\sqrt{pq}, \qquad \frac1{pr}=\frac1{\sqrt{pq}}.$$
Hence
$$\theta_{\max} = \arctan\sqrt{pq} - \arctan\frac1{\sqrt{pq}}.$$
For positive $x$,
$$\arctan x+\arctan\frac1x=\frac\pi2.$$
Applying this with $x=\sqrt{pq}$,
$$\theta_{\max} = 2\arctan\sqrt{pq}-\frac\pi2.$$
For
$$p=2,\qquad q=\frac32,$$
we have
$$pq=3,$$
so
$$\theta_{\max} = 2\arctan\sqrt3-\frac\pi2 = 2\cdot\frac\pi3-\frac\pi2 = \frac\pi6.$$
Thus
$$\boxed{\frac{|AB|}{|AD|}=\sqrt{\frac q p}}$$
and
$$\boxed{\max\angle PAQ = 2\arctan\sqrt{pq}-\frac\pi2}.$$
In the particular case $p=2$, $q=\frac32$,
$$\boxed{\max\angle PAQ=\frac\pi6},$$
with equality when
$$\boxed{\frac{|AB|}{|AD|}=\sqrt{\frac34}=\frac{\sqrt3}{2}}.$$
Verification of Key Steps
The first delicate step is the formula
$$\theta = \arctan\frac q r-\arctan\frac1{pr}.$$
The direction angle of $\overrightarrow{AP}$ is $\arctan!\left(\frac1{pr}\right)$, while the direction angle of $\overrightarrow{AQ}$ is $\arctan!\left(\frac q r\right)$. Since $q>1$, the second slope is larger than the first:
$$\frac q r-\frac1{pr} = \frac{pq-1}{pr}>0.$$
Hence the angle between the rays is exactly the difference of these direction angles. Forgetting to verify this inequality could produce an incorrect sign.
The second delicate step is solving the critical point equation. From
$$p(r^2+q^2)=q(p^2r^2+1)$$
one obtains
$$p(1-pq)r^2=q(1-pq).$$
The factor $1-pq$ may be cancelled only after checking that it is nonzero. Since $p,q>1$, $pq>1$, so the cancellation is legitimate and yields
$$r^2=\frac q p.$$
The third delicate step is proving maximality. Rewriting the derivative as
$$\theta'(r) = \frac{(pq-1)(q-pr^2)} {(r^2+q^2)(p^2r^2+1)}$$
shows that its sign is determined solely by $q-pr^2$. This gives a positive sign before the critical point and a negative sign after it, establishing a global maximum rather than merely a stationary point.
Alternative Approaches
A purely geometric approach starts from the identity
$$\tan\angle PAQ = \frac{\tan\alpha-\tan\beta} {1+\tan\alpha\tan\beta},$$
where
$$\alpha=\arctan\frac q r,\qquad \beta=\arctan\frac1{pr}.$$
Substitution gives
$$\tan\angle PAQ = \frac{pq-1}{pqr+r^{-1}}.$$
Since $pq-1$ is constant, maximizing the angle is equivalent to maximizing its tangent, hence to minimizing
$$pqr+\frac1r.$$
The arithmetic-geometric mean inequality yields
$$pqr+\frac1r \ge 2\sqrt{pq},$$
with equality exactly when
$$pqr=\frac1r,$$
that is,
$$r^2=\frac q p.$$
This method avoids differentiation and reaches the optimum through a single inequality. The coordinate and calculus approach is preferable because it also produces the value of the maximal angle directly and makes the monotonicity completely transparent.