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A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.

This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.

Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).

Thanks in advance.

  • $\begingroup$ If you revisit how the ellipse equation is derived, Pyhthagoras and trig are involved .So the ellipse equation usage is an indirect usage combined shortcut with ellipse equation .. $2 c= 2, c=1; 2a =4, a=2; b^2=a^2-c^2= =3$ $ \text{ Plug into ellipse equation } \dfrac{(y+1)^2}{2^2}+ \dfrac{1^2}{3}=1;\to y+1 \to \sqrt\frac{8}{3};$ Or so it appears to me. $\endgroup$
    – Narasimham
    Jan 3, 2021 at 18:17

6 Answers 6


This is a slight variation on the answer of @leoli1 which omits the square roots. So start with $$\begin{eqnarray} y^2+1&=& \lvert BC \rvert^2 \\ (y+2)^2+1&=& \lvert AC \rvert^2 \end{eqnarray}$$ but now note instead that $$ \lvert AC \rvert^2 = (4 - \lvert BC \rvert)^2.$$ Subtract the first equality above from the second to find $$4y + 4 = 16 - 8 \lvert BC \rvert$$ so $\lvert BC \rvert = (3 - y)/2$. Substitute this back in the first equality to get a quadratic equation for $y$.

  • 3
    $\begingroup$ Oh yeah, this is a better way, it will save us a lot of tedious calculations. $\endgroup$
    – leoli1
    Jan 3, 2021 at 9:06

Notice that by Pythagoras we have: \begin{align} y^2+1&= \left\vert BC\right\vert^2\\ (2+y)^2+1&= \left\vert AC\right\vert ^2 \end{align} Taking roots and adding both equations yield: $$\sqrt{y^2+1}+\sqrt{(2+y)^2+1}=\left\vert BC\right\vert+\left\vert AC\right\vert=4$$ Now we can solve this equation to get the value of $y$.

  • $\begingroup$ That should give a quartic equation, after removing squares; it doesn't look simple to solve (at least with pen and paper). $\endgroup$ Jan 2, 2021 at 18:36
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    $\begingroup$ @FedericoPoloni actually the cubic and quartic term will cancel, so we are left with a quadratic equation. $\endgroup$
    – leoli1
    Jan 2, 2021 at 18:53
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    $\begingroup$ I don't think that squaring an equation with two surds, removing the $x^2$ terms, rearranging, and squaring again is the sort of algebra that high school students would be comfortable with. I stared at that equation for minutes before seeing no other way and hoping I'm not getting a quartic. $\endgroup$
    – obscurans
    Jan 3, 2021 at 1:35
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    $\begingroup$ @silverfish oh I agree. To me at least, staring at the problem, I immediately write the natural-looking equation that expresses the lengths of the string. It's the algebra of solving it that's a hairy mess. To wit, I think path of least resistance (in the beginning) will lead ~most people towards this solution. $\endgroup$
    – obscurans
    Jan 3, 2021 at 2:46
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    $\begingroup$ @Countour-Integral: you've omitted the (y^2 +1) term. $\endgroup$
    – smci
    Jan 3, 2021 at 8:58

I will give a simple solution using ellipse which is rather more easier.

Since the thread is tied to $A$ and $B$, we deduce that $A$ and $B$ are the focii of the ellipse. If we rotate the ellipse in the counter clockwise direction and make it horizontal, then we can use the equation for standard ellipse which is $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ where $AD$ now becomes the $x$ axis, $DC$ the positive $y$ axis, with origin being the midpoint of $AB$. So difference between focii is given to be $2ae=2$ and length of string $=2a=4$ which implies $a=2$ and $e=\dfrac{1}{2}$. So, $b^2=a^2(1-e^2)=4(1-0.25)=3$. So, the equation of ellipse becomes, $$\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$$ Putting $y=1$, we get $x=\pm \sqrt{\dfrac{8}{3}}$. So, $BD=\sqrt{\dfrac{8}3}-1$.

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    $\begingroup$ I said in the question I don't want to use ellipses. $\endgroup$ Jan 2, 2021 at 10:45
  • $\begingroup$ Because I'm a maths tutor and I tutor students who haven't been introduced to ellipses yet. $\endgroup$ Jan 2, 2021 at 10:53
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    $\begingroup$ @AdamRubinson, I think it's alright to just have an answer by this approach as well, maybe other readers will find it helpful, a solution using pythagoras is already written so I hope your query would have been already answered. $\endgroup$
    – V.G
    Jan 2, 2021 at 14:51
  • $\begingroup$ It’s fine. However just because there is a solution using pythagoras doesn’t mean that’s the * only* elementary approach. No one has given a solution with trigonometric functions for example, so if it exists then that would be even better. $\endgroup$ Jan 2, 2021 at 15:26

We know the side $c=2$ and the altitude $h_c=1$, hence the area of the triangle is $S=\frac12ch_c=1$.

We also have Heron's formula for the area:


With $p=\dfrac{a+b+c}2=3$.





Now, we know $a+b$ and $ab$, hence:


And since the larger side is $b$,



Now, with Pythagoras' theorem in the triangle BDC:


  • 1
    $\begingroup$ @AdamRubinson I understand your concern. However, Heron's formula is not very difficult at secondary school level (you can prove it with the law of cosines as is shown on Wikipedia). And probably not more difficult than a quadratic equation. $\endgroup$ Jan 2, 2021 at 12:18
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    $\begingroup$ @AdamRubinson In many school curriculums, Heron's formula is taught at secondary level. Usually after Pythagoras, sometimes at a similar time to introducing sin/cos/tan for right-angled triangles, and generally not later than introducing rule of sines or cosines for non-right-angled triangles. If taught at all, almost certainly done so long before R addition formulas - in fact many school systems will teach ellipses before that! Asking on the world-wide web for a solution involving "elementary" rather than "advanced" techniques often shows how arbitrary and variable teaching order is! $\endgroup$
    – Silverfish
    Jan 2, 2021 at 21:26
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    $\begingroup$ Related to that, I'm a big fan of the law of tangents and law of cotangents, which often provide very neat solutions to trig problems. $\endgroup$
    – Silverfish
    Jan 2, 2021 at 21:27
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    $\begingroup$ If you needed to use trig tables and pen-and-paper calculation, fewer steps was a big advantage, so the laws of tangents and cotangents were widely taught at secondary level. The sine and cosine laws might get you there in a few more steps, but with electronic calculators to do the hard work, this hardly matters: hence most curricula have dropped the laws of tangent and cotangent, most schoolkids wouldn't have a clue about them, and yet the laws have not become more "advanced", just more neglected! So as Jean-Claude says, "advanced" turns out to be quite arbitrary :-) $\endgroup$
    – Silverfish
    Jan 2, 2021 at 21:33
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    $\begingroup$ @Silverfish Not only more steps, but also precision loss. With a slide rule or a table of logs, one must take great care of precision, and the law of cosines is terrible, for small angles. I learnt trigonometry on in my father's schoolbooks (published in the early 1960s), and they are great for that. $\endgroup$ Jan 2, 2021 at 21:42

This is the closest I could come up with to a purely "trigonometric" answer, and has minimal algebra. Firstly note you have a triangle divided into two, so it may be useful to use the standard formulae for a divided triangle. Quoting from my (rather old!) secondary school mathematics textbook:

When D divides AB in the ratio $m:n$ then: $$(m+n) \cot \theta = n \cot A - m \cot B = m \cot \alpha - n \cot \beta \tag{1}$$ triangle ABC where D divides AB

Here the large triangle is $\triangle ABC$ and the Greek-lettered angles are $\alpha = \angle ACD$, $\beta = \angle BCD$ and $\theta = \angle BDC$ and is supplementary to $\angle ADC$. Given your example is about a string attached to a wall, you might not be surprised this is listed in the "Mechanics" section of the textbook, and usually used in school examinations to solve three-force problems. There is a very useful special case listed, which sadly doesn't apply here but I'll include it for completeness:

When $D$ is the midpoint of $AB$, then $$2 \cot \theta = \cot A - \cot B = \cot \alpha - \cot \beta $$

The textbook has $B$ and $D$ labelled the other way round to your diagram, but other than that it just takes a rotation to sort out. We have $m = 2$, $n = y$, $BC = 1$ and $B$ (using the textbook formula's labels) is a right angle with $\cot B = 0$.

From $\triangle BCD$ we have $\cot \theta = \frac{y}{1} = y$ and $\cot \beta = \frac{1}{y}$, while $\triangle ABC$ gives $\cot A = \frac{2+y}{1} = 2 + y$. Hence $(1)$ becomes:

$$(2 + y) \cdot y = y \cdot (2 + y) - 2 \cdot 0 = 2 \cot \alpha - y \cdot \frac{1}{y}$$

The first equality doesn't tell us anything new, but the second yields (with a little bit of algebra):

$$y^2 + 2y = 2 \cot \alpha - 1 \\ \implies (y+1)^2 = 2 \cot \alpha \tag{2}$$

Secondly, note that since $AC + BC = 4$, we know that $\triangle ABC$ has perimeter 6 and semi-perimeter 3. Hence it is very inviting to use Heron's formula or the closely-related law of cotangents. In fact, since for the one side we know, we also have the altitude (perpendicular height) and hence can calculate the area, this temptation becomes overwhelming. I find the most practically useful formulation of the law of cotangents when solving a triangle is:

$$\frac{\cot \frac{\alpha}{2}}{s-a} = \frac{1}{r} = \frac{s}{T} \tag{3}$$

or for calculator purposes, should we lack a cot button, the reciprocal:

$$(s - a) \tan \frac{\alpha}{2} = r = \frac{T}{s}$$

Here $\alpha$ and $a$ are any opposite angle-side pair (indeed the formula is often stated with equalities also for similar expressions involving $\beta$ and $\gamma$, resembling the law of sines, but I find that form is generally far more useful for trigonometric proof than solving a given triangle), $r$ is the radius of the incircle (not relevant here but sometimes useful), $T$ is the area of the triangle and $s$ the semi-perimeter. In our case $(3)$ becomes:

$$ \frac{\cot \frac{\alpha}{2}}{3-2} = \frac{3}{ \frac{1}{2} \cdot 2 \cdot 1 } \\ \implies \cot \frac{\alpha}{2} = 3 \tag{4} $$

With an electronic calculator, it is now straightforward to calculate $\alpha$ and hence $2 \cot \alpha$, substitute into $(2)$ and solve the simple quadratic for $y$. If an exact or non-calculator solution is required, we can instead use the double-angle formula:

$$ \cot \alpha = \frac{\cot^2 \frac{\alpha}{2} - 1}{2 \cot \frac{\alpha}{2}} = \frac {3^2 - 1}{2 \cdot 3} = \frac{4}{3}$$

Then the positive solution of $(y+1)^2 = \frac{8}{3}$ is

$$y = \sqrt{\frac{8}{3}} - 1 = \frac{2\sqrt{6}}{3} - 1 \approx 0.63$$

and corresponds to the bead's vertical level lying beneath the lower attachment point, as drawn. By symmetry, the bead will also be a horizontal distance of one metre from the wall when it is held this far above the vertical level of the upper attachment point. In terms of vertical distance above the lower attachment point, this is then around $2.63$ metres, which — as we might hope — coincides neatly with the absolute value of the negative root of the quadratic.

Ignore my verbal commentary, and this method gets to the answer in just a couple of lines of working, involves no algebraic rearrangement except solving a simple quadratic, and packs a very hearty dose of trigonometry from a high school textbook. Accept the use of an electronic calculator after $(2)$ and it doesn't even require the angle-sum, double-angle or harmonic addition ("R") formulae. Sadly, I expect it is the case that your local schools' curriculum lacks some of the goodies in my textbook. This is a shame! The first time I saw the divided triangle formula, I recall the sinking feeling that it looked like an awful lot of hard work. What I've discovered in practice, whenever I've had recourse to use it, is just how much hard work it saves me...

Post-script: proof of the divided triangle formulae

Many proofs are available online and in textbooks for the law of cotangents but the divided triangle formula seems more obscure.

Often half the battle with trig is identifying the correct formula to use: which best represents the state of our knowledge? For example, knowing the area, perimeter and one side of a triangle, the law of cotangents is ideal for finding the angle opposite the known side. Here's another rule true for any $\triangle PQR$ and deduced by dropping the perpendicular from $Q$, which a little thought shows is true in both the acute and obtuse case:

Triangle PQR with perpendicular QF dropped from Q to PR

$$q = r \cos P + p \cos R \tag{5}$$

where $q = PR$ etc. Formula $(5)$ is useful if we want to write an included side in terms of its including angles and the other two sides. In practice it's rare that we are given four out of five of these quantities, or wish to represent all five in one equation. However, $p$, $r$, $P$ and $R$ are redundant by the law of sines, so we can cut one out:

\begin{align} q &= r \cos P + \frac{r \sin P}{\sin R} \cdot \cos R \\ &= r \left( \cos P + \sin P \cot R \right) \\ &= r \sin P \left( \frac{\cos P}{\sin P} + \cot R \right) \\ q &= r \sin P \left( \cot P + \cot R \right) \tag{6} \end{align}

Formula $(6)$ lets us write an included side in terms of its including angles and one other side. Like the law of sines, this formula links two sides and two angles. Unlike the law of sines, there is only one opposite angle-side pair!

Returning to the figure in the textbook, the key to representing the situation is that the smaller triangles $\triangle ACD$ and $\triangle BCD$ are linked by a common side $CD$ and the fact $ADB$ is a straight line, so $\angle ADC$ and $\angle BDC$ are supplementary. Hence $CD$, $\angle ADC$ and $\angle BDC$ must appear in our equations if those facts are to be brought to bear. Moreover, we want $AD = m$ and $BD = n$ to appear too, since they were indicated as of particular interest (the ratio $\triangle ABC$ was divided in for the textbook; the lengths marked on the wall in the OP's question). Needing to represent two sides and an included angle in each of the smaller triangles rules out the law of sines; we could apply the law of cosines if we wanted $AC$ and $BC$ to appear in our equations also. Alternatively we can apply $(4)$ if we prefer two sides and two angles in each equation.

Highlighted version of first figure

At this stage we have a choice. Treating $AD$ as included by $A$ and $\pi - \theta$, and $BD$ as included by $B$ and $\theta$,

\begin{align} m &= CD \sin (\pi - \theta) \left(\cot (\pi - \theta) + \cot A \right) = CD \sin \theta \left( -\cot \theta + \cot A \right) \\ n &= CD \sin \theta \left( \cot \theta + \cot B \right) \\ \implies \frac{m}{n} &= \frac{-\cot \theta + \cot A }{\cot \theta + \cot B} \tag{7} \end{align}

where $(5)$ is obtained by dividing the previous two equations. Alternatively we can treat $CD$ as included by $\alpha$ and $\pi - \theta$ in $\triangle ACD$ and included by $\beta$ and $\theta$ in $\triangle BCD$:

\begin{align} CD &= m \sin (\pi - \theta) \left(\cot (\pi - \theta) + \cot \alpha \right) = CD \sin \theta \left( -\cot \theta + \cot \alpha \right) \\ CD &= n \sin \theta \left( \cot \theta + \cot \beta \right) \\ \implies 1 &= \frac{m}{n} \cdot \frac{-\cot \theta + \cot \alpha }{\cot \theta + \cot \beta} \tag{8} \end{align}

Rearranging $(7)$ and $(8)$ to combine them into the textbook's $(1)$ is straightforward algebra.


Here's an alternative trigonometric approach that doesn't assume knowledge of the divided triangle formulae, but at the expense of additional work algebraically.

ABD vertical, DC horizontal

Half the battle with trigonometry is getting started: which formulae capture the key details of your set-up? If we want to do trigonometry separately on the smaller triangles $\triangle ABC$ and $\triangle BCD$, the following points are essential:

  • $ABD$ is a straight line, so $\angle ABC$ and $\angle CBD$ are supplementary,
  • The two triangles share the length $BC = x$.

So we expect to need these angles and sides in our equations. For $\triangle ABC$ we also know $AB = 2$ and $AC = 4 - x$. A way to incorporate all three sides and $\angle ABC$ would be the law of cosines:

\begin{align} \cos(\angle ABC ) &= \frac{2^2 + x^2 - (4-x)^2}{2 \cdot 2 \cdot x} \\ \implies -\cos(\angle CBD ) &= \frac{8x - 12}{4x} \\ \implies \cos(\angle CBD) &= \frac{3}{x} - 2 \tag{1} \\ \end{align}

where we used the fact $\angle ABC$ and $\angle CBD$ are supplementary, because we want to relate this finding to the other triangle.

Since $\triangle BCD$ is right-angled, we can stick to basic "SOHCAHTOA" rather than anything fancy, and as discussed above we certainly want to make use of $BC = x$ and $\angle CBD$. There are two choices which both give the solution, though in very different ways.

Take the sine of $\angle CBD$; combine with cosine by harmonic addition

Since $\sin (\angle CBD) = \frac{1}{x}$ then $(1)$ implies that

$$ \cos \theta - 3 \sin \theta = -2$$

This can be solved by the harmonic addition ("R") formula to deduce an acute angle of approximately 57.7°; using $\tan (\angle CBD) = \frac{1}{y}$ we have

$$y = \frac{1}{\tan (\angle CBD) } \approx 0.63$$

Note that the harmonic addition formula also produces an obtuse solution around 159.2°, for which we obtain $y \approx -2.63$, corresponding to the bead lying 0.63 vertically above $A$ instead of 0.63 vertically below $B$.

Take the cosine of $\angle CBD$; compare results and form a quadratic

From $\triangle ABC$ we had deduced $\cos(\angle CBD) = \frac{3}{x} - 2$.

From $\triangle BCD$ we have $\cos(\angle CBD) = \frac{y}{x}$.

Comparing results we have:

\begin{align} \frac{y}{x} &= \frac{3}{x} - 2 \\ \implies y &= 3 - 2x \end{align}

Filling that into our diagram we see $AD = y+2 = 5-2x$. We now have all the lengths of the right-angled triangle $\triangle ADC$ in terms of $x$, so we can form a quadratic using Pythagoras. Note we could instead write the sides in terms of $y$, but there would be fractional coefficients which make unnecessarily hard work. It's easier to find $x$ then use $y = 3-2x$ to convert back to $y$ at the end.

\begin{align} (5 - 2x)^2 + 1^2 &= (4-x)^2 \\ \implies 3x^2 -12x + 10 &= 0\\ \implies 3(x-2)^2 &= 2 \\ \implies x = 2 \pm \sqrt {\frac 2 3} &= 2 \pm \frac{\sqrt 6 }{3} \end{align}

Note that it's the lower of these two values which yields the positive root for $y$,

$$y = 3 - 2x = -1 + \frac{2\sqrt 6 }{3} \approx 0.63$$

while the negative root

$$y= -1 - \frac{2\sqrt 6 }{3} \approx -2.63$$

corresponds to the solution above $A$, as before.

Alternative: $\angle BAC = \angle DAC$ is common to $\triangle ABC$ and $\triangle ACD$

This works like the above, but has the slight advantage it doesn't require students to know $\cos (\pi - \theta) = - \cos (\theta)$. It makes use of the fact $ABD$ is a straight line by looking at how a small triangle lies inside the larger one. Using the law of cosines in $\triangle ABC$:

\begin{align} \cos(\angle BAC) &= \frac{2^2 + (4-x)^2 - x^2}{2 \cdot 2 \cdot (4-x)} \\ \implies \cos(\angle BAC) &= \frac{20 - 8x}{4(4-x)} \\ \implies \cos(\angle BAC) &= \frac{5-2x}{4-x} \tag{2} \end{align}

Then in the right-angled triangle $\triangle ACD$, we have

\begin{align} \cos (\angle BAC) &= \frac{y+2}{4-x} \tag{3} \\ \sin (\angle BAC) &= \frac{1}{4-x} \tag{4} \end{align}

We can use $(2)$ and $(3)$ together to write $y$ in terms of $x$.

\begin{align} \frac{5-2x}{4-x} &= \frac{y+2}{4-x} \\ \implies 5-2x &= y + 2 \\ \implies y = 3 - 2x \end{align}

We can apply Pythagoras to $\triangle ACD$ and solve for $x$ and $y$, as above.

Alternatively, we can use $(2)$ and $(4)$ together to relate the sine and cosine:

\begin{align} (2) \implies \cos(\angle BAC) &= \frac{5-2x}{4-x} \\ &= \frac{8-2x-3}{4-x} \\ &= 2 - 3 \cdot \frac{1}{4-x}\\ (4) \implies \cos(\angle BAC) &= 2 - 3 \sin (\angle BAC) \\ \implies \cos(\angle BAC) + 3 \sin (\angle BAC) &= 2 \end{align}

Which can then be solved e.g. by harmonic addition to produce an acute angle of 20.8° and obtuse angle of 122.3°. We can then solve for $y$ e.g. by using $\tan(\angle BAC) = \frac{1}{2+y}$.

  • $\begingroup$ Almost certainly not the "best" way to do it! Just wanted to submit an alternative that shows a primarily trigonometric approach is possible, while using only well-known formulae. $\endgroup$
    – Silverfish
    Jan 3, 2021 at 18:19

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