# Simpler solution to this geometry/trig problem?

i had a geometry/trignometry problem come up at work today, and i've been out of school too long: i've lost my tools.

i'm starting with a rectangle of known width (w) and height (h). For graphical simplification i can convert it into a right-angle triangle:

i'm trying to find the coordinates of that point above which is perpendicular to the origin:

i've labelled the opposite angle t1 (i.e. theta1, but Microsoft Paint cannot easily do greek and subscripts), and i deduce that the two triangles are similar (i.e. they have the same shape):

Now we come to my problem. Given w and h, find x and y.

Now things get very difficult to keep drawing graphically, to explain my attempts so far.

But if i call the length of the line segment common to both triangles M:

then:

M = w∙sin(t1)


Now i can focus on the other triangle, which i'll call O-x-M:

and use trig to break it down, giving:

x = M∙sin(t1)
= w∙sin(t1)∙sin(t1)

y = M∙cos(t1)
= w∙sin(t1)∙cos(t1)


with

t1 = atan(h/w)


Now this all works (i think, i've not actually tested it yet), and i'll be giving it to a computer, so speed isn't horribly important.

But my god, there must have been an easier way to get there. i feel like i'm missing something.

By the way, what this will be used for is drawing a linear gradient in along that perpendicular:

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I think your approach is very straightforward ... it's certainly correct! If you're concerned that your final formulas look overly-complicated, it's because you stopped too soon. To finish up, you should eliminate the references to $t_1$. If you happen to know how to simplify "the sine (or cosine) of the arctan of $t_1$", great. If not, you may notice that the sine and cosine appear in your picture. Writing $s$ for the length of the hypotenuse, you have $\sin t_1 = h/s$ and $\cos t_1 = w/s$. Thus, your formulas reduce to $x=w h^2/s^2$ and $y=hw^2/s^2$. Then, since $s^2 = x^2 + y^2$ ... – Blue Jan 19 '11 at 11:49
Re: Eliminating $t_1$ ... Notice that, with your formulas, you'd compute $t_1$ via an inverse-trig function, only to toss the result into trig functions. Inverse-trigs turn "length info" into "angle info", while trigs turn "angle info" into "length info". Your solution passes through "angle info" on a journey from "length info" back to "length info". When this happens, it's a good bet that there's an algebraic expression that makes the journey directly. Here, you should suspect that $\cos {\rm atan} h/w$ is expressible algebraically in terms of $h/w$; and it is: $1/\sqrt{1+(h/w)^2}$ (or $w/s$) – Blue Jan 19 '11 at 12:18
It might looks more straightforward because i've already traveled the route, but it took me 45 minutes to get there. While i was getting there i thought, this can't be right, there has to be a better way. i had to find the size of a 3rd implied triangle, based on the similarity of two previous triangles. And sin(cos(atan)), "Oh boy, this is getting nasty." – Ian Boyd Jan 19 '11 at 12:49

An equation for the hypotenuse of the right triangle is $y=-\frac{h}{w}x+h$. Its slope is $-\frac{h}{w}$ (change in $y$-coordinate over change in $x$-coordinate). The slope of any line perpendicular to the hypotenuse is $\frac{w}{h}$ (perpendicular lines have slopes with product $-1$, or it is sometimes said that the slope of a perpendicular line is the "opposite reciprocal"), so an equation for the line through the origin and perpendicular to the hypotenuse is $y=\frac{w}{h}x$. The point you want is the solution to the system of equations: \begin{align}y&=-\frac{h}{w}x+h\\y&=\frac{w}{h}x\end{align}

These equations give two expressions that are both equal to $y$, so set them equal: $$-\frac{h}{w}x+h=\frac{w}{h}x$$ and solve:$$h=\left(\frac{w}{h}+\frac{h}{w}\right)x=\frac{w^2+h^2}{wh}x$$ $$x=\frac{h^2w}{w^2+h^2}$$ Use one of the original equations to find $y$: $$y=\frac{w}{h}x=\frac{hw^2}{w^2+h^2}$$

edit Working from Hans Lundmark's answer, final method, a vector from the origin to the desired point is in the same direction as $\langle\frac{1}{w},\frac{1}{h}\rangle$ or equivalently $\langle h,w\rangle$. The distance from the origin to the desired point is the length of the altitude—call this $d$. Looking at the area of the triangle, it is both $\frac{1}{2}wh$ and $\frac{1}{2}d\sqrt{w^2+h^2}$, so $d=\frac{wh}{\sqrt{w^2+h^2}}$. The desired vector is thus \begin{align} \frac{\langle h,w\rangle}{|\langle h,w\rangle|}d &=\frac{\langle h,w\rangle}{\sqrt{w^2+h^2}}\frac{wh}{\sqrt{w^2+h^2}} \\ &=\frac{\langle h,w\rangle wh}{w^2+h^2} \\ &=\left\langle\frac{h^2w}{w^2+h^2},\frac{w^2h}{w^2+h^2}\right\rangle .\end{align}

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I just want to make a plug for the "intercept-intercept form" of the line equation: $\frac{x}{w}+\frac{y}{h}=1$ :) – Blue Jan 19 '11 at 11:38
Of the two answers that provide final form, this is the one i understand more :) Accepted. – Ian Boyd Jan 19 '11 at 14:16

For fun, here's another approach ...

The area of your triangle is given by $\frac{1}{2} w h$, but it's also given by $\frac{1}{2}s m$, where $s$ is the length of the hypotenuse and $m$ is the length of your segment $M$. (To see why, flip the triangle over so that it's sitting on its hypotenuse; then, clearly, $s$ is the "base" and $m$ the "height".) Therefore,

$$w h = s m$$

Writing $\theta$ for the angle that segment $M$ makes with the horizontal, we have that

$$x = m \cos\theta = \frac{wh}{s}\cos\theta \hspace{0.5in} y = m\sin\theta = \frac{wh}{s}\sin\theta$$

By similar triangles, $\theta$ is also the angle in the top corner of your triangle, so

$$\cos\theta = \frac{h}{s} \hspace{0.5in} \sin\theta = \frac{w}{s}$$

and we have

$$x = \frac{wh^2}{s^2}=\frac{w h^2}{w^2+h^2} \hspace{0.5in} y = \frac{w^2 h}{s^2}=\frac{w^2h}{w^2+h^2}$$

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Parametrize the line from $(w,0)$ to $(0,h)$ by $(w,0) + t(-w, h)$. Then you are searching for the point $(x,y)$ on the line such that $(x,y)\cdot (-w,h) = 0$.

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Is the "*" in the final equation intended to be dot-product? – Isaac Jan 19 '11 at 1:58
Yes, that is the dot product. – M.B. Jan 19 '11 at 15:12
I've edited your answer to render the ordered pairs, expression, and equation using $\LaTeX$. Feel free to re-edit or revert if you don't like the result. – Isaac Jan 19 '11 at 21:18

Another way: You have identified some angles as being equal in your last figure. This implies that several of the right triangles in the picture are similar. For example, $$\frac{h}{w} = \frac{x}{y} = \frac{y}{w-x},$$ from which it is not too difficult to solve for $x$ and $y$.

And yet another: The hypotenuse lies on the line with equation $x/w+y/h=1$ (since the two points $(x,y)=(w,0)$ and $(x,y)=(0,h)$ satisfy this equation, and two points uniquely determine a line). The normal vector to a line can be read off from the coefficients of $x$ and $y$; it is $\mathbf{n}=(1/w,1/h)$ in this case. The point that you seek (let me call it $(a,b)$ here) lies on the line which goes from the origin in the direction that $\mathbf{n}$ points, so the point must be of the form $(a,b)=t \mathbf{n} = (t/w,t/h)$ for some number $t$. Substituting $(x,y)=(t/w,t/h)$ into the equation for the hypotenuse gives $t(1/w^2+1/h^2)=1$, from which we immediately find $t$ and hence also $$(a,b)=\left( \frac{1/w}{1/w^2+1/h^2}, \frac{1/h}{1/w^2+1/h^2} \right).$$

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Suppose $h = (h_1,h_2)$ and $w = (w_1, w_2)$. Then you know the slope of the line. Let's call this $m_1$. You also know that the line perpendicular to the point $(x,y)$ has slope $-\frac{1}{m_1}$. You get a system of equations.

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