Maximize Area of A Rectangle.

Question

I have a rectangle $ABCD$ and $P$ is a point inside the rectangle and the distance from the point $P$ to all the vertices of the rectangle is given. Now I have to figure out the maximum possible area of the rectangle.

For example: If $PA=13$ , $PC=47$ , $PD=43$ , $PB = 23$ , then what can be the maximum area of the rectangle?

I know single-variable calculus , but here the main problem is that I cannot get an equation involving only one variable.
Two variables are coming into question.

I have also tried it using pure geometry by dropping perpendiculars from the point P to the other sides, then applying Pythagoras but doesn't help. Also, angle chasing is not the case here I guess.

Answer 1

We have the Diagram like this:

$ABCD$ is the rectangle, with Point $P$ inside.

The unknown Segments are $w,x,y,z$.

Pythagoras Theorem: We have the following Equations:

$w^2+x^2=13^2 \tag{1}$
$x^2+z^2=23^2 \tag{2}$
$y^2+z^2=47^2 \tag{3}$
$y^2+w^2=43^2 \tag{4}$

Solve the Set of Equations to get the values. The rectangle will have sides $z+w$ and $x+y$. The Area is then Obtained.

We get: $x^2 = 169 - w^2, y^2 = 1849 - w^2, z = w^2 + 360$, all in terms of $w$.

$$x = \sqrt{169 - w^2}, y = \sqrt{1849 - w^2}, z = \sqrt{w^2 + 360}$$

The Area is:
$$(z+w)(x+y)=[\sqrt{w^2 + 360}+w][\sqrt{169 - w^2}+\sqrt{1849 - w^2}]$$

We have a function in 1 variable ($w$), hence we can differentiate & set that to $0$ to get the $w$ value for local Maximum. Plug that value to get the other unknowns and then get the Maximum Area.

Answer 2

I will give two solutions, the elegant one is the geometric one. The reader in hurry may please want to skip to (3) for a very short geometric proof.

Why am i starting differently? The idea for the geometric construction arises after having the dirty solution, the algebraic one. So both solutions have their merit. I will plot them in the order i found them, the first one will be the algebraic no-insight solution. Having it, we try to find a geometric interpretation. This is easily done, we are even constrained to construct as we construct. When the natural geometric elements are there, the geometric solution shows up immediately.

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Let us solve the problem in full generality.

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(1) First (algebraic) solution and discussion:

We are giving four numbers $a,b,c,d>0$. We are searching for a geometrical configuration which contains

• a rectangle $ABCD$ with sides $p,q$, taken without restriction to have $A=(0,0)$, $B=(p,0)$, $C=(p,q)$, $D=(0,q)$,
• and a point $\Pi=(x,y)$ (inside or outside $ABCD$) having distances $a,b,c,d$ to respectively $A,B,C,D$,

such that $pq$ is maximal among all such possible configurations.

First of all, let us collect algebraically the constraints on the variables $p,q,x,y$: $$ \tag{$*$} $$ $$ \left\{ \begin{aligned} x^2 + y^2 &= a^2\ ,\\ (p-x)^2 + y^2 &= b^2\ ,\\ (p-x)^2 + (q-y)^2 &= c^2\ ,\\ x^2 + (q-y)^2 &= d^2\ . \end{aligned} \right. $$ We observe that from the above we have $$ \tag{$**$} a^2 + c^2= x^2+y^2+(p-x)^2+(q-y)^2= b^2 + d^2\ . $$ So there is a condition to be satisfied by the input $a,b,c,d$, namely $a^2+c^2=b^2+d^2$. (And this is the case in the given numerical case: $13^2 + 47^2 = 2378 =43^2 + 23^2$.)

This condition is also geometrically easy to understand. Project $P$ on the four sides, the projection points are $P_{AB}$, $P_{BC}$, $P_{CD}$, $P_{DA}$, we obtain four smaller rectangles with opposite corners $(P,A)$, respectively $(P,B)$, $(P,C)$, $(P,D)$. Using the theorem of Pythagoras to express the squared diagonals $a^2, b^2,c^2, d^2$ of them in terms of the squared projections on the sides, we immediately see that $a^2+c^2=b^2+d^2$.

So we have three equations in four unknowns. There is "one free direction". We need to find a maximum w.r.t. this degree of freedom for the expression $pq$. So let us restate the problem.

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Problem: Let $a,b,c,d$ be four positive numbers that satisfy $a^2+c^2=b^2+d^2$. Denote the common value by $e^2$. Consider the set of all $x,y\in\Bbb R$ and all $p,q>0$ so that the relations $(*)$ above are satisfied. Which is the maximal value of the product $pq$?

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Short (algebraic) answer: $pq\le ac+bd$.

So in the given case it is $$\ (13\cdot 47 + 23\cdot 43)=\bbox[lightgreen]{\ \bf 1600\ }\ . $$ The equality is obtained as in the following picture. And the green area $ \bbox[lightgreen]{\bf 1600}$ in the picture was shown as value for the rectangle, it is the computed geogebra area. The picture is "involved", because the numbers $13, 23, 43, 47$ were not so easy to transport from the axes into their final places.

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This is, as stated in the Problem a purely algebraic problem. It is useful to introduce the following notations, (so $A,B,C,D$ are in this algebraic section no longer the vertices of a rectangle) that make life easier:

We denote by capitalized letter variables the squares of lower case variables.

So now $A,B,C,D;P,Q;X,Y$ stay for $a^2, b^2,c^2,d^2;p^2,q^2;x^2,y^2$ respectively. Let us solve the algebraic problem.

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Solution for the algebraic problem:

First of all we eliminate $x,y$. From the first two equations in $(*)$, subtracting them, we obtain:

$P+A-B=p^2+a^2-b^2=2px$. Similarly for the middle two equations

$Q+B-C=q^2 + b^2 -c^2 =2qy$.

We extract from the linear dependencies $x,y$, and plug them in into the first equation. This gives successively: $$ \begin{aligned} 4PQ\; A &= 4p^2q^2a^2 = 4p^2q^2x^2+ 4p^2q^2y^2 \\ &=Q(2px)^2 + P(2qy)^2 =Q(P+A-B)^2 + P(Q+B-C)^2\ , \\[3mm] &\text{ In simpler form:} \\[3mm] 0 &= PQ(P+Q) -2PQ(A+C) + Q(A-B)^2 + P(B-C)^2\ . \end{aligned} $$ We equivalently want to maximize $PQ$ constrained to the above condition, in the domain $P,Q>0$. It is clear that the maximum is not reached at the boundary, which is $PQ=0$. (Because the value at the boundary of the function to be maximized is $PQ=0$.) So the maximal value is also a local extremal value, it is thus obtained as a critical point for the (Lagrange multiplicators) help function $h$ (and i use $m$ instead of $\lambda$ for an easy typing): $$ h(P,Q;m):=PQ-m\Bigg(\ PQ(P+Q) -2PQ(A+C) + Q(A-B)^2 + P(B-C)^2\ \Bigg)\ . $$ Such a critical point satifies the three equations $0=h'_P=h'_Q=h'_m$, and the resulted algebraic system is: $$ \left\{ \begin{aligned} 0 &= h'_Q = P - m\Big(\ P(2Q+P) - 2P(A+C) + (A-B)^2 \ \Big)\ ,\\ 0 &= h'_P = Q - m\Big(\ Q(2P+Q) -2Q(A+C) + (B-C)^2\ \Big) \ ,\\ 0 &= -h'_m = PQ(P+Q) -2PQ(A+C) + Q(A-B)^2 + P(B-C)^2\ . \end{aligned} \right. $$ We eliminate $m$ from the first two equations above, $$ \frac {PQ}m = PQ(2Q+P) -2PQ(A+C)+Q(A-B)^2\ ,\\ \frac {PQ}m = PQ(2P+Q) -2PQ(A+C)+P(B-C)^2\ , $$ and subtract: $$ 0 = PQ(P-Q) - Q(A-B)^2 + P(B-C)^2\ . $$ Add now the above to the third relation $0=h'_m$ to obtain the simpler equation, since $Q(A-B)^2$ cancels: $$ P^2Q - PQ(A+C) + P(B-C)^2\ . $$ We can factorize $P\ne0$. We finally get $$ \tag{$\dagger$} PQ=Q(A+C)-(B-C)^2\ . $$ A similar equation is obtained by adding instead of subtracting above, and making that $P(B-C)^2$ cancels. We factorize $Q\ne0$. We obtain a linear relation (both sides being equal to $PQ$): $$ Q(A+C)-(B-C)^2 = P(A+C) - (A-B)^2\ . $$ We extract $P$ as a function of $Q$, $P=Q +(A-C)(D-B)/(A+C)$, and plug in into $(\dagger)$. The resulted equation of second degree in $Q$ is: $$ \tag{$\dagger\dagger$} 0 = Q^2 + \frac 1{A+C}\Big(\ (A-C)(D-B) - (A+C)^2\ \Big)Q + (B-C)^2\ . $$ The coefficient of $Q$ involves $(A-C)(D-B) - (A+C)^2 =(A-C)(D-B) - (A+C)(D+B) =-2(AB+CD)$. The half discriminant is $$ \left(\frac{AB+CD}{A+C}\right)^2-(B-C)^2 =\frac 1{A+C}^2\Big(\ (AB+CD)^2 -(B-C)^2(A+C)^2\ \Big) =\frac 1{A+C}^2\Big(\ (AB+CD)^2 -(AB-CD)^2\ \Big) =\frac {4ABCD}{A+C}^2=\frac{4a^2b^2c^2d^2}{A+C}^2\ . $$ So we can write the possible solutions: $$ (q^2)_\pm Q_\pm= \frac{AB+CD}{A+C}\pm \frac{2abcd}{A+C}\ . =\frac 1{A+C}{a^2b^2+c^2d^2\pm 2abcd} =\frac {(ab+cd)^2}{a^2+c^2}\ . $$ And we take of course only the bigger value as a critical value for reaching the maximum. (There is only one choice, and we take it.) $$ q_+ =\frac{ab+cd}{\sqrt{a^2+c^2}}\ . $$ In a similar manner, the point $p_+$ for the solution of the system, and thus the only one relevant as a critical value is: $$ p_+ =\frac{ad+bc}{\sqrt{a^2+c^2}}\ . $$ So the maximal value of the product is: $$ \bbox[yellow]{\qquad p_+q_+ =\frac 1{a^2+c^2}(ab+cd)(ad+bc) =ac+bd \qquad\ .\ } $$ $\square$

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It looks like all the above computations are painful. Well, there is an alternative to get the same result (checked) in only few lines, below in sage.

R.<a,b,c,d,p,q,m> = PolynomialRing(QQ)
A, B, C, D, P, Q = a^2, b^2, c^2, d^2, p^2, q^2
h = P*Q - m*( -4*P*Q*A + Q*(P + A - B)^2 + P*(Q + B - C)^2 )
J = R.ideal([ diff(h, P), diff(h, Q), diff(h, m), A + C - B - D])
gen = J.elimination_ideal([m, d, p]).gens()[0]
print(factor(gen))

And there are three easy factors, and a "complicated" one... Manually rearranged output:

(-a + b)^2 * (a + b)^2 * q^4 
* (a^2*b^4 - 2*a^2*b^2*c^2 + b^4*c^2 + a^2*c^4 - 2*b^2*c^4
       + c^6 - 2*a^2*b^2*q^2 - 2*a^2*c^2*q^2 + 2*b^2*c^2*q^2
       - 2*c^4*q^2 + a^2*q^4 + c^2*q^4)

The complicated one, expr below, is giving the equation of degree two in $Q=q^2$ from $(\dagger\dagger)$. Using now the computer to solve needs some few more lines.

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(2) Geometric interpretation:

Geometric problem: Let $a,b,c,d>0$ be so that $a^2+c^2=b^2+d^2$, and denote the common value by $e^2$. (We may and do assume $a\le b\le d\le c$, so that the picture corresponds to the one below.) Then the numbers $a,c,e$ and $b,d,e$ are the sides of two right triangles, and we draw them inscribed in a circle on both sides (half-planes) of a diameter with length $e$: Use points $\Psi,\Pi;S,T,T'$ labeled as in the picture. Then the biggest area of a rectangle that accepts a point with distances $a,b,c,d$ to its vertices is: $$ac+bd=2[S\Psi T\Pi]\ ,$$ and the rectangle realizing this maximum has sides $f,F$.

Indeed, using Ptolemy we have $ab+cd=eF$ and $ad+bc=ef$. So in the above notations $q_+=(ab+cd)/e=eF/e=F$, and $p_+ = (ad+bc)/e=ef/e=f$.

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(3) Proof for the geometric version:

(The picture is the proof. Try to figure it out here, without reading further.)

In the picture we start with the rectangle $ABCD$ and a point $\Pi$ having distances $a,b,c,d$ to its distances. Construct now points $P_{AB}$, $P_{BC}$, $P_{CD}$, $P_{DA}$ as in the picture. The point $\Pi_{DA}$ is obtained for instance by copy+pasting the triangle $\Delta \Pi BC$ after a translation in direction $\vec{BA}=\vec{CD}$. Similar constructions apply for the other points. After all copy+paste steps the area $[ABCD]$ is doubled. We can estimate geometrically: $$ \begin{aligned} 2[ABCD] &= [P_{AB}\; B\; P_{BC}\; C\; P_{CD}\; D\; P_{DA}\; A] \\ &= \sum_\text{cyclic} [\Pi A\Pi_{AB}B] = \sum_\text{cyclic} \frac 12 \Pi\Pi_{AB}\cdot AB \\ &\le \sum_\text{cyclic} \frac 12 (ac+bd)=2(ac+bd)\ . \end{aligned} $$ The equality is obtained when $\Pi A\Pi_{AB}B$ (and/or the other three quadrilaterals) are cyclic, we use Ptolemy.

$\square$

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(4) Construction of the maximizing configuration: Start with the figure in (2). Recall that $ST=f$, $ST'=F$. Let $U$ be opposite to $S$ in the circle $(S\Psi TT'\Pi)$. SO $SU$ diameter, and $\widehat{ST'U}=90^\circ$. Construct perpendiculars in $S,T'$ on $ST'$ and on them segments of length $f$. We obtain a rectangle $ST'XY$ as in the figure.

The points $T,Y$ are mirrored w.r.t. $S\Pi$, for instance following the path

$ \widehat{YS\Pi} = 90^\circ - \widehat{T'S\Pi} = 90^\circ - \widehat{T'\Psi\Pi} = \widehat{\Psi\Pi T'} = \widehat{\Pi\Psi T} = \widehat{\Pi ST} $.

So $\Pi Y=\Pi T=d$. It follows also $\Pi X=a$, because $\Pi$ has the lengths $b,c,d$ to three vertices of a rectangle.

The area of $[ST'XY]$ is $fF=p_+q_+=ac+bd$, the maximal value.

$\square$