A problem about perpendicular lines and areas bounded by coordinate axes

Question

I did not want to publish this problem and I wanted to solve it myself, but I was tired of dealing with this difficult problem. This problem, in a way, tells us that we need some theorems that start from areas to calculate lengths. There are many theorems in the opposite direction, but I hardly see any theorems that start from areas. The lengths are calculated

The problem first occurred to me four days ago

the problem:

We have the Descarte coordinate system and two perpendicular straight lines intersecting at a point that does not belong to the coordinate axes and does not match the coordinate axes, and they limit the Areas $A,B,C$ shown in the above figure. If the data are $A,B,C$ values, how Can we find out the coordinates of their intersection $P$? (Suppose we know at what level of the Cartesian level the point $P$) there are several ways I tried to deal with the problem but I haven't come up with a final solution yet, I will put some ideas on the problem:

perhaps the first idea and the easiest idea is to know that the two triangles are in the area. The same is equal to the square of the similarity, this idea gave me a large number of proportions that can be formulated, and certainly the theorem of Pythagoras also appears with these existing triangles and square proportions, but this method was a great confusion, in front of me a large number Of the equations and distractions and I don't know what to do.

A second idea to proceed with the solution: we know that one of the definitions of the hiperbola is that it represents the geometrical place of a moving straight line that confines a fixed space by its intersection with two fixed intersections.

$f(x):y=\frac{-(A+C)}{2x}$ , $g(x):y=\frac{B}{2x}$

$f'(x)=\frac{A+C}{2x^2}$ , $g'(x)=\frac{-B}{2x^2}$

$\frac{A+C}{2{x_1}^2}\cdot\frac{-B}{2{x_2}^2}=-1$

$x_2=\frac{\sqrt{B(A+C)}}{2x_1}$

$Δ_1:y+\frac{A+C}{2x_1}=\frac{A+C}{2{x_1}^2}\cdot(x-x_1)$

$Δ_2:y-\frac{B}{2x_2}=\frac{-2{x_1}^2}{A+C}\cdot(x-x_2)$

But I stopped here and couldn't finish

I suppose we can continue by finding the equation of the area enclosed between $Δ_1,Δ_2$ and the two coordinate axes and then finding when this equation equals $A$

A third way to deal with the problem is to calculate the lengths in the figure in terms of an optional major length and then deal with that length separately

$PS=a$

$OS=\frac{a\sqrt{A+C}}{\sqrt{A+B}}$

$OT=\frac{2B\sqrt{A+B}}{a\sqrt{B}}$

$PT=\frac{2(A+B)}{a}$

$OV=\frac{a\sqrt{B}}{\sqrt{A+B}}$

All of these methods may give some hope and insight to solve this specific problem, but I think that the problem is that the problem is in the lack of theorems that start from the Areas and give a few lengths, and we must start making and collecting them. The lengths are based on given areas.

Thank you

Answer 1

Assuming that $P = (x_1, y_1) $ where $x_1 \lt 0$ and $y_1 \gt 0 $, as shown in the given figure. Also, it is assumed that the points $U, S, V, T$ have a position similar to what is shown in the given figure.

Let $\theta = \angle SUO $, an unknown angle.

Then the line $US$ has the equation

$- \sin \theta (x - x_1) + \cos \theta (y - y_1) = 0 $

It follows that

$U = (U_x, 0) = ( x_1 + \cot \theta (-y_1) , 0 ) $

And,

$S = (0,S_y) = (0, y_1 + \tan \theta \ (-x_1) ) $

Hence,

$\begin{align}A + C &= \dfrac{1}{2} | U_x S_y | = \dfrac{1}{2} | (x_1 - \cot \theta y_1)(y_1 - \tan \theta x_1) | \\ &= \dfrac{1}{2} ( x_1^2 \tan \theta + y_1^2 \cot \theta - 2 x_1 y_1 ) \end{align}$

$V, T$ are on $PT$ whose equation is

$ \cos \theta (x - x_1) + \sin \theta (y - y_1) = 0$

therefore,

$V = ( x_1 - \tan \theta (-y_1) , 0 )$

$T = (0, y_1 - \cot \theta (-x_1) )$

From the figure,

$A + B = \dfrac{1}{2} | x_1 (S_y - T_y) | = \dfrac{1}{2} x_1^2 ( \tan \theta + \cot \theta )$

And finally,

$ C = 1/2 | y1 | | Vx - Ux | = \dfrac{1}{2} y_1^2 ( \tan \theta + \cot \theta )$

From the last two equations,

$\dfrac{C}{A + B} = \left(\dfrac{y_1}{x_1}\right)^2 = t^2$

where $t \lt 0$ because $P$ is in the second quadrant.

So that

$ y_1 = t x_1 \hspace{10pt}(1)$

Substitute this into the first equation,

$A + C = \dfrac{1}{2 \sin \theta \cos \theta} x_1^2 (\sin^2 \theta + t^2 \cos^2 \theta - 2 t \sin \theta \cos \theta )$

but

$x_1^2 = \dfrac{2(A + B)}{ \tan \theta + \cot \theta} = 2 (A + B) \sin \theta \cos \theta \hspace{10pt}(2)$

Therefore,

$\dfrac{A + C}{A + B} = \sin^2 \theta + t^2 \cos^2 \theta - 2 \ t \ \sin \theta \cos \theta $

And this simplifies to,

$\dfrac{A + C}{A + B} = \dfrac{1}{2}(1 + t^2) + \dfrac{1}{2}(t^2 - 1) \cos(2 \theta) - t \sin(2 \theta) $

This equation is straight forward to solve, because it is of the form

$ c_1 \cos(\phi) + c_2 \sin(\phi) = c_3 $

where

$\begin{align}\phi &= 2 \theta \\ c_1 &= \dfrac{1}{2} (t^2 - 1) \\ c_2 &= - t \\ c_3 &= \dfrac{A + C}{A + B} - \dfrac{1}{2}(1 + t^2) \end{align}$

Its solution is

$ \phi = 2 \theta = \alpha \pm \beta $

where

$ \alpha = \text{ATAN2}( c_1, c_2 )$

$ \beta = \cos^{-1}\left( \dfrac{c_3}{\sqrt{c_1^2 + c_2^2} } \right) $

(The $\text{ATAN2}(c_1, c_2)$ function is a standard function in computer languages and returns the angle $\alpha$, where $\cos \alpha = \dfrac{c_1}{\sqrt{c_1^2 + c_2^2} } $ and $\sin \alpha = \dfrac{c_2}{\sqrt{c_1^2 + c_2^2}} $)

Now we have $\theta$, so we can compute $x_1$ from equation $(2)$, and then we can compute $y_1$ from equation $(1)$.

Example:

I'll cook an example, to show exactly how the above equations work.

So moving in the forward direction, suppose we pick $P = (-4, 1) $

And suppose we choose the line $US$ to be

$ y = \dfrac{1}{2} \ x + 3 $

Clearly $P$ is on this line. The perpendicular line $(VT)$ passing through $P$ is

$ y = -2 x - 7$

From the above two equations

$ U = (-6, 0) , S = (0, 3) , V = ( - \dfrac{7}{2} , 0 ) , T = (0, -7)$

It follows that

$ A = \dfrac{1}{2} ( (4)(2) + (4 + \dfrac{7}{2} ) (1) ) = \dfrac{31}{4} $

$ B = \dfrac{1}{2} ( \dfrac{7}{2} ) (7) = \dfrac{49}{4} $

$ C = \dfrac{1}{2} (1)( - \dfrac{7}{2} + 6 ) = \dfrac{5}{4} $

Now, we'll take the values of $A,B,C$ and use the developed method above to find the coordinates of $P$.

First thing, calculate $t$

$ t^2 = \dfrac{C}{A + B} = \dfrac{5}{80} = \dfrac{1}{16} $

So

$ t = - \dfrac{1}{4} $

Next, calculate $c_1 ,c_2 , c_3$.

$ c_1 = -\dfrac{15}{32} $

$ c_2 = \dfrac{1}{4} $

$ c_3 = \dfrac{36}{80} - \dfrac{1}{2} (\dfrac{17}{16} ) = \dfrac{9}{20} - \dfrac{17}{32} = -\dfrac{13}{160}$

So the equation we want to solve, after scaling is

$ - 75 \cos(\phi) + 40 \sin(\phi) = - 13 $

And we get following the outlined method,

$ \alpha = \text{ATAN2}(-75, 40) = 2.65163532734 $

$ \beta = \cos^{-1}\left( \dfrac{-13}{\sqrt{75^2 + 40^2} } \right) = 1.72434010933 $

And we get two values for $\theta$

$\theta = \dfrac{1}{2} (\alpha + \beta) = 2.1879 $ (This value is extraneous, because it does not lie in the first quadrant)

OR

$ \theta = \dfrac{1}{2} ( \alpha - \beta ) = 0.463647609 $

One can check that $\tan(\theta) = \dfrac{1}{2} $ as expected.

Now, from equation $(2)$

$ x_1 = - \sqrt{ 2 (A + B) \sin \theta \cos \theta } = - \sqrt{ 2 (20) \dfrac{2}{5} } = -4 $

And from equation $(1)$

$ y_1 = t x_1 = - \dfrac{1}{4} (-4) = 1 $

Hence, $ P = (-4, 1) $

Here is a listing of a small program to implement these calculations.

Public Sub calculating_P()

Dim a, b, c As Double
Dim t, c1, c2, c3 As Double
Dim alpha, beta, theta As Double
Dim x1, y1 As Double

a = ActiveSheet.Cells(1, 1)
b = ActiveSheet.Cells(1, 2)
c = ActiveSheet.Cells(1, 3)

' Compute t
t = -Sqr(c / (a + b))

' compute theta

c1 = 0.5 * (t ^ 2 - 1)
c2 = -t
c3 = (a + c) / (a + b) - 0.5 * (1 + t ^ 2)

alpha = WorksheetFunction.Atan2(c1, c2)
beta = WorksheetFunction.Acos(c3 / Sqr(c1 ^ 2 + c2 ^ 2))

theta = 0.5 * (alpha + beta)

If theta > 0 And theta < p / 2 Then

x1 = -Sqr(2 * (a + b) * Sin(theta) * Cos(theta))
y1 = t * x1

ActiveSheet.Cells(3, 1) = x1
ActiveSheet.Cells(3, 2) = y1

ActiveSheet.Cells(3, 5) = theta
End If

theta = 0.5 * (alpha - beta)

If theta > 0 And theta < p / 2 Then

x1 = -Sqr(2 * (a + b) * Sin(theta) * Cos(theta))
y1 = t * x1

ActiveSheet.Cells(4, 1) = x1
ActiveSheet.Cells(4, 2) = y1

ActiveSheet.Cells(4, 5) = theta
End If



End Sub

And below is the Input/Output of the program:

Answer 2

I'll start with a bit of an extended comment, to address OP's thought that

This problem, in a way, tells us that we need some theorems that start from areas to calculate lengths.

Consider an acute triangle $\triangle ABC$ with altitudes from $A$ and $B$ meeting their opposing sides at $D$ and $E$, and meeting each other at orthocenter $H$, through which the altitude from $C$ also passes. (Assuming the triangle is acute ensures that $H$ lies in its interior. An analogous analysis for obtuse triangles is left as an exercise to the reader.)

Define $$P := |\triangle AEH| \qquad Q := |\triangle BDH| \qquad R := |\square CDHE|$$ The essence of OP's coordinate-finding task is determining $\triangle ABC$ from areas $P$, $Q$, $R$. So, let's concentrate on that.

First, note that we can calculate the area of $\square ACBH$ thusly: $$P+Q+R = \frac12 |CH|\,c \tag{1}$$ It happens that we can write $P$ and $Q$ individually in terms of $|CH|$ as well ...

Slide $H$ parallel to $AC$ to a point $H_A$ that completes parallelogram $\square ECHH_A$; since $\triangle AEH_A$ has the same base ($AE$) and corresponding height as $\triangle AEH$, we see that $|\triangle AEH_A|=P$. But $\triangle AEH_A$ can also be considered to have "base" $|EH_A|$ (which is equal to $|CH|$); the corresponding altitude is readily determined to be $c\cos^2 A$, so that we have $$P = \frac12 |CH|\,c\cos^2 A \qquad\text{likewise,}\qquad Q = \frac12 |CH|\,c\cos^2 B \tag2$$ Consequently, $$\cos^2 A = \frac{P}{P+Q+R} \qquad \cos^2 B = \frac{Q}{P+Q+R}$$ These relations determine (acute) angles $A$ and $B$, which in turn determine $C$. The shape of $\triangle ABC$ is now known.

Now, let $r$ is the triangle's circumradius. By the (Extended) Law of Sines and an elementary property of the orthocenter, we have
$$\left.\begin{array}{r} c = 2r\sin C \\ |CH| = 2r\cos C \end{array}\right\} \quad\to\quad P+Q+R=\frac12\cdot 2r\cos C\cdot 2r\sin C = r^2\sin2C\tag{3}$$ Thus, we determine $r$, so that the scale of $\triangle ABC$ is known.

We have therefore completely recovered $\triangle ABC$, as desired. $\square$

---

To complete OP's coordinate-finding task, we re-orient $\triangle ABC$ appropriately, and note that the coordinates of $E$ are given by $$\begin{align} x &= -a\cos C\sin C = \,-2 r \sin A\cos C\sin C = -r\sin2C\sin A \\ y &= \phantom{-}c\cos A\cos C = \phantom{-}2r\sin C\cos C\cos A = \phantom{-}r\sin2C\cos A \end{align}$$ By the above, $$\begin{align} \cos A &= \sqrt{\frac{P}{P+Q+R}} \qquad \sin A = \sqrt{1-\cos^2A} = \sqrt{\frac{Q+R}{P+Q+R}} \\[8pt] r\sin2C &= \sqrt{r^2\sin^22C} = \sqrt{(P+Q+R)\sin2C} \end{align}$$ and $$\begin{align} \sin2C &= 2\sin C\cos C = -2\sin(A+B)\cos(A+B) \\ &= 2(\sin A\cos B+\cos A\sin B)( \sin A\sin B-\cos A\cos B ) \\ &= \frac{2}{(P+Q+R)^2} \left((-P + Q + R) \sqrt{Q (P + R)} + (P - Q + R) \sqrt{P (Q + R)} \right) \\ \end{align}$$ That is,

$$\begin{align} x &= -\frac{\sqrt{2S(Q+R)}}{P+Q+R} \qquad y = \frac{\sqrt{2SP}}{P+Q+R} \\[8pt] S &:= (-P + Q + R) \sqrt{Q (P + R)} + (P - Q + R) \sqrt{P (Q + R)} \end{align}$$

Demonstrating that this is (or isn't!) equivalent to results in other answers is left as another exercise to the reader. However, using the numerical example from @Hosam's answer, $$P = 5/4 \qquad Q = 49/4 \qquad R = 31/4$$ we have $$P+Q+R = 85/4 \qquad Q+R = 20\qquad S = 1445/8$$ $$\sqrt{2S(Q+R)} = 85 \qquad \sqrt{2SP} = 85/4$$ $$(x,y) = (-4,1)$$ which agrees with Hosam's calculation. This doesn't seem like a coincidence.

Answer 3

This is a solution in Arabic, I don't have the patience to rewrite it in English, it contains a lot of equations so I hope it is understandable

Answer 4

TL;DR. Closed forms for the coordinates of $P$, given areas $A, B, C$:

$$x_P = -\frac{\sqrt{2} (A + B)}{\Omega \left[ \sqrt{A + C} + \frac{1}{A} \sqrt{B} \left( \sqrt{(A + C) B} + \sqrt{(A + B) C} \right) \right]}$$

$$y_P = \frac{\sqrt{2} C}{\Omega \left[ \frac{1}{A} \sqrt{A + C} \left( \sqrt{(A + C) B} + \sqrt{(A + B) C} \right) - \sqrt{B} \right]}$$

where

$$\Omega = \sqrt{\frac{\sqrt{(A + C) B} - \sqrt{(A + B) C}}{B - C}}$$

If it is the case that $B > C$ and $\sqrt{B} < \frac{1}{A} \sqrt{A + C} \left( \sqrt{(A + C) B} - \sqrt{(A + B) C} \right)$, there is an additional solution:

$$x_P = -\frac{\sqrt{2} (A + B)}{\Omega' \left[ \sqrt{A + C} + \frac{1}{A} \sqrt{B} \left( \sqrt{(A + C) B} - \sqrt{(A + B) C} \right) \right]}$$

$$y_P = \frac{\sqrt{2} C}{\Omega' \left[ \frac{1}{A} \sqrt{A + C} \left( \sqrt{(A + C) B} - \sqrt{(A + B) C} \right) - \sqrt{B} \right]}$$

where

$$\Omega' = \sqrt{\frac{\sqrt{(A + C) B} + \sqrt{(A + B) C}}{B - C}}$$

I don't know whether these expressions can be simplified much, unfortunately, but you can verify that they are correct.

Denote the line $\overleftrightarrow{VT}$ by $L_1$ and the line $\overleftrightarrow{US}$ by $L_2$.

Here is how you can derive these. As I mentioned in the comments, the key is to use similarity. Let $x > 0$ denote $OV$, the distance between the origin and the $x$-intercept of $L_1$ (do not confuse $x$ with $x_P$). Let $y > 0$ denote $OS$, the distance between the origin and the $y$-intercept of $L_2$ (do not confuse $y$ with $y_P$). Finally, let $-k$ be the slope of $L_1$ (note that $k > 0$ and the slope of $L_2$ is $\frac{1}{k}$). The variable $k$ encodes part of the similarity we will exploit.

With these definitions, we have the following system:

$$\begin{cases} 2 B = k x^2 \\ 2 (A + C) = k y^2 \\ \left( \frac{x \sqrt{1 + k^2}}{y + k x} \right)^2 = \frac{B}{A + B} \end{cases}$$

where the first equation is obtained by the area of $\triangle VTO$, the second equation is obtained by the area of $\triangle SUO$, and the third equation expresses the fact that the ratio of areas of $\triangle VTO \sim \triangle STP$ is equal to the square of the ratio of the hypotenuses $VT$ and $ST$.

Solving this system, we get

$$k = \frac{1}{A} \left( \sqrt{(A + C) B} \pm \sqrt{(A + B) C} \right)$$

Now, we need $k > 0$. In general, we can take the "plus" root. If it is further the case that $B > C$, we can also consider the "minus" root. For the "plus" root, we get that

$$\begin{cases} x = \sqrt{2 B} \Omega \\ y = \sqrt{2 (A + C)} \Omega \end{cases}$$

where $\Omega$ is defined above. In the same way, for the "minus" root, we get that

$$\begin{cases} x = \sqrt{2 B} \Omega' \\ y = \sqrt{2 (A + C)} \Omega' \end{cases}$$

where $\Omega'$ is defined above. Of course, to ensure that $y > 0$ in the case of the "minus" root, we enforce

$$\sqrt{B} < \frac{1}{A} \sqrt{A + C} \left( \sqrt{(A + C) B} - \sqrt{(A + B) C} \right)$$

To conclude, simply use the areas of $\triangle STP$ and $\triangle VUP$, respectively, to find

$$\begin{cases} x_P = -\frac{2 (A + B)}{y + k x} \\ y_P = \frac{2 C}{k y - x} \end{cases}$$

Answer 5

Calling

$$ \cases{ p=(x,y)\\ p_0 = (x_0,y_0)\\ R(\theta) = \left(\matrix{\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta}\right)\\ T(\theta,p,p_0) = R(\theta)\cdot p+p_0 } $$

the coordinates of $U,V,S,T$ are calculated according to

$$ \cases{ T(\theta,(0,y'),P)=(0,y), \ \ \to S\\ T(\theta,(0,y'),P)=(x,0), \ \ \to U\\ T(\theta,(x',0),P)=(x,0), \ \ \to V\\ T(\theta,(x',0),P)=(0,y), \ \ \to T } $$

once we have $\{U,V,T,S\}$ as functions of $\{P,\theta\}$ we calculate the areas concerning the respective perimeters

$$ \cases{ \sum\{O,S,P,V,O\}=\sigma_A(P,\theta) = A\\ \sum\{V,T,O,V\}= \sigma_B(P,\theta) = B\\ \sum\{P,U,V,P\}= \sigma_C(P,\theta) = C } $$

Three equations for three unknowns. Follows a MATHEMATICA script performing those calculations.

parms = {aB -> 0.1, aC -> .1, aA -> 1};
Trans[p_, t_, p0_] := RotationMatrix[t].p + p0
area[p_] := Module[{xy, X, Y}, xy = Transpose[p];
  X = xy[[1]];
  Y = xy[[2]];
  Return[Sum[(Y[[k + 1]] + Y[[k]]) (X[[k + 1]] - X[[k]]), {k, 1, Length[X] - 1}]/2]]
pP = {x0, y0};
pO = {0, 0};
solS = Solve[Trans[{0, ys}, theta, pP] - {0, y} == 0, {ys, y}][[1]];
pS = {0, y} /. solS;
solU = Solve[Trans[{0, ys}, theta, pP] - {x, 0} == 0, {x, ys}][[1]];
pU = {x, 0} /. solU;
solV = Solve[Trans[{xs, 0}, theta, pP] - {x, 0} == 0, {xs, x}][[1]];
pV = {x, 0} /. solV;
 solT = Solve[Trans[{xs, 0}, theta, pP] - {0, y} == 0, {xs, y}][[1]];
pT = {0, y} /. solT;
perimC = {pP, pU, pV, pP};
perimB = {pV, pT, pO, pV};
perimA = {pO, pS, pP, pV, pO};
areaA = area[perimA]
areaB = area[perimB]
areaC = area[perimC]
equs = {areaA == aA, areaB == aB, areaC == aC} /. parms;
solsA = Quiet@Solve[equs, {x0, y0, theta}, Reals] /. {C[1] -> 0}//N
perim = {pS, pU, pV, pT, pS};
perims = Table[perim /. solsA[[k]], {k, 1, 4}]
gr1 = ListPlot[perims[[1]], PlotStyle -> {PointSize[0.02], Blue}];
gr1b = ListLinePlot[perims[[1]]];
gr1a = Graphics[{Black, PointSize[0.02], Point[pP /. solsA[[1]]]}];
Show[gr1, gr1a, gr1b,PlotRange -> {{-2.5, 0.5}, {-2.5, 0.5}}, AspectRatio -> 1]

Follows one of the solutions for $A=1,B=0.1, C=0.1$.