Who wouldn't attempt to solve it after that commentary? Poor spider, though. That must have been tiring.

I'm certain any readers would want to attempt this for themselves as well, so my solution (and the accompanying story!) can be found after the jump break.

Spoilers ahead!

The approach I took was to name six unknown variables, then build a system of nonlinear equations which could be reduced using substitution. Why six? I tried fewer, but realized that this produced an underdetermined problem. I named these variables as follows:

Observe the addition of a red dividing line with the same length as $x$. From here, the properties of transversal lines and vertical angles can be applied to determine which angles are congruent to each other:

which allows us to identify several similar triangles (highlighted in similar colors, some are congruent):

(Goodness! The colors looked much less garish when I was putting it together.)

We may now build the initial system of equations. Since $a$ and $c$ together comprise the full length of the 12-foot string, $a+c=12$; and similarly, $b+d = 10$. We use the Pythagorean Theorem to find that the top portion of the red bar has length $\sqrt{a^2-e^2}$ and the bottom portion has length $\sqrt{b^2-f^2}$. Both of these sides belong to two triangles each, so, by using the Pythagorean Theorem and transitive property on the shared side, we obtain $b^2-f^2 = c^2-25$ and $a^2-e^2=d^2-25$.

Similar triangle properties can be used to find the last two equations. Due to conservation of ratios in similar triangles, $\frac{a}{e} = \frac{c}{5}$, $\frac{b}{f}=\frac{d}{5}$ and $\frac{b}{d} = \frac{c}{a}$. We may now focus on reducing the system of all stated equations, that is,

\begin{align}

a+c&=12\quad &(1)\\\\

b+d&=10\quad &(2)\\\\

b^2-f^2 &= c^2-25\quad &(3)\\\\

a^2-e^2 &=d^2-25\quad &(4)\\\\

\frac{a}{e} &= \frac{c}{5}\quad &(5)\\\\

\frac{b}{f}&=\frac{d}{5}\quad &(6)\\\\

\frac{b}{d} &= \frac{c}{a}\quad &(7).

\end{align}

If you are expecting the next, more difficult, step of the problem to be an unintelligible mass of algebra... you are absolutely correct.

Let's begin by using (1), (2) and (7) to derive an important relationship. Noting that (7) can be rewritten as $ab=cd$, we can solve (1) for $b$ and (2) for $c$ to make the substitution

$$a(10-d)=(12-a)d,$$

which simplifies to $d = \frac{5}{6}a$. This new information can be used to modify (4) to

$$a^2-e^2=\frac{25}{36}a^2-25.$$

Finding this last equation allows us to separate a system of two equations with two variables, namely

\begin{align*}

\frac{a}{e} &= \frac{c}{5}\\\\

a^2-e^2&=\frac{25}{36}a^2-25.

\end{align*}

From the second we solve for $e=\frac{5a}{12-a}$, hence the last equation becomes

\begin{align}

a^2-\left(\frac{5a}{12-a}\right)^2&=\frac{25}{36}a^2-25\quad\textrm{or}\\\\

11 a^4-264 a^3+1584 a^2-21600 a +129600&= 0

\end{align}

(mod annoying algebra.) This polynomial has two imaginary and two real solutions; we reject the imaginary solutions and consider the real ones $a\approx 6.645$, $a\approx 20.26$. (Anyone is welcome to put the following polynomial into Wolfram Alpha to see why I'm not presenting exact solutions.) As the second solution is larger than the size of the string, we reject it and conclude $a=6.645$.

Now the above equations determine

\begin{align}

c &= 12 - a = 5.355,\\\\

d &= \frac{5}{6}a = 5.538,\\\\

b &= 10 - d = 4.462,\\\\

e &= \frac{5a}{c} = 6.204,\\\\

f &= \frac{5b}{d} = 4.029.

\end{align}

We have arrived at the final step: recall that $x$ is the sum of the two red lines in the diagram, hence

$$x = \sqrt{a^2-e^2}+\sqrt{b^2-f^2}=4.298.$$

Hooray!

There's also an interesting story behind how I did this problem. Some of you who have read this blog before may remember that I am engaged to a very brilliant man---a man who clearly surpasses me in mathematical skill in problems that aren't directly related to my field, and who has had a much easier time being accepted by academia. I ask him for this reason to

Yet this time he did, and solved the problem in about 10 minutes... while I toiled at the whiteboard for over an hour.

I became very angry with myself. I stared at the equations until my eyes hurt. I tried several combinations of equations only to find that some were dependent in odd ways. After what could only be described as a math-induced breakdown, I asked my partner, how? How did he solve this in only ten minutes?

That's when he looked at the diagrams he drew, and the equations I was working with, and paused. "I may have made an inconsistent assumption," he stated.

We determined that he decided to simplify the problem by assuming the angles at the intersection of the two strings were all right angles, which makes the algebra nicer, but doesn't satisfy the constraints of the problem (the wall would have to be larger on the left side to accommodate the strings). I had avoided making this assumption and was able to reduce the system to a fourth-order polynomial in one variable soon after.

I tell this story because it highlights the importance of teamwork in math: even if one person is clearly the smartest, problem solving is ultimately about creativity, and having more people thinking about a problem only means that 'aha!' moments/mistake catches will occur faster. This is a fantastic mindset to have if you, like myself, often spend time interacting with much smarter people.

As there may be a more elegant solution, I'd be happy to hear about the approach you took in the comments!

I'm certain any readers would want to attempt this for themselves as well, so my solution (and the accompanying story!) can be found after the jump break.

Spoilers ahead!

The approach I took was to name six unknown variables, then build a system of nonlinear equations which could be reduced using substitution. Why six? I tried fewer, but realized that this produced an underdetermined problem. I named these variables as follows:

Observe the addition of a red dividing line with the same length as $x$. From here, the properties of transversal lines and vertical angles can be applied to determine which angles are congruent to each other:

which allows us to identify several similar triangles (highlighted in similar colors, some are congruent):

(Goodness! The colors looked much less garish when I was putting it together.)

We may now build the initial system of equations. Since $a$ and $c$ together comprise the full length of the 12-foot string, $a+c=12$; and similarly, $b+d = 10$. We use the Pythagorean Theorem to find that the top portion of the red bar has length $\sqrt{a^2-e^2}$ and the bottom portion has length $\sqrt{b^2-f^2}$. Both of these sides belong to two triangles each, so, by using the Pythagorean Theorem and transitive property on the shared side, we obtain $b^2-f^2 = c^2-25$ and $a^2-e^2=d^2-25$.

Similar triangle properties can be used to find the last two equations. Due to conservation of ratios in similar triangles, $\frac{a}{e} = \frac{c}{5}$, $\frac{b}{f}=\frac{d}{5}$ and $\frac{b}{d} = \frac{c}{a}$. We may now focus on reducing the system of all stated equations, that is,

\begin{align}

a+c&=12\quad &(1)\\\\

b+d&=10\quad &(2)\\\\

b^2-f^2 &= c^2-25\quad &(3)\\\\

a^2-e^2 &=d^2-25\quad &(4)\\\\

\frac{a}{e} &= \frac{c}{5}\quad &(5)\\\\

\frac{b}{f}&=\frac{d}{5}\quad &(6)\\\\

\frac{b}{d} &= \frac{c}{a}\quad &(7).

\end{align}

If you are expecting the next, more difficult, step of the problem to be an unintelligible mass of algebra... you are absolutely correct.

Let's begin by using (1), (2) and (7) to derive an important relationship. Noting that (7) can be rewritten as $ab=cd$, we can solve (1) for $b$ and (2) for $c$ to make the substitution

$$a(10-d)=(12-a)d,$$

which simplifies to $d = \frac{5}{6}a$. This new information can be used to modify (4) to

$$a^2-e^2=\frac{25}{36}a^2-25.$$

Finding this last equation allows us to separate a system of two equations with two variables, namely

\begin{align*}

\frac{a}{e} &= \frac{c}{5}\\\\

a^2-e^2&=\frac{25}{36}a^2-25.

\end{align*}

From the second we solve for $e=\frac{5a}{12-a}$, hence the last equation becomes

\begin{align}

a^2-\left(\frac{5a}{12-a}\right)^2&=\frac{25}{36}a^2-25\quad\textrm{or}\\\\

11 a^4-264 a^3+1584 a^2-21600 a +129600&= 0

\end{align}

(mod annoying algebra.) This polynomial has two imaginary and two real solutions; we reject the imaginary solutions and consider the real ones $a\approx 6.645$, $a\approx 20.26$. (Anyone is welcome to put the following polynomial into Wolfram Alpha to see why I'm not presenting exact solutions.) As the second solution is larger than the size of the string, we reject it and conclude $a=6.645$.

Now the above equations determine

\begin{align}

c &= 12 - a = 5.355,\\\\

d &= \frac{5}{6}a = 5.538,\\\\

b &= 10 - d = 4.462,\\\\

e &= \frac{5a}{c} = 6.204,\\\\

f &= \frac{5b}{d} = 4.029.

\end{align}

We have arrived at the final step: recall that $x$ is the sum of the two red lines in the diagram, hence

$$x = \sqrt{a^2-e^2}+\sqrt{b^2-f^2}=4.298.$$

Hooray!

There's also an interesting story behind how I did this problem. Some of you who have read this blog before may remember that I am engaged to a very brilliant man---a man who clearly surpasses me in mathematical skill in problems that aren't directly related to my field, and who has had a much easier time being accepted by academia. I ask him for this reason to

*not*look over my shoulder when I'm trying to solve a problem.Yet this time he did, and solved the problem in about 10 minutes... while I toiled at the whiteboard for over an hour.

I became very angry with myself. I stared at the equations until my eyes hurt. I tried several combinations of equations only to find that some were dependent in odd ways. After what could only be described as a math-induced breakdown, I asked my partner, how? How did he solve this in only ten minutes?

That's when he looked at the diagrams he drew, and the equations I was working with, and paused. "I may have made an inconsistent assumption," he stated.

We determined that he decided to simplify the problem by assuming the angles at the intersection of the two strings were all right angles, which makes the algebra nicer, but doesn't satisfy the constraints of the problem (the wall would have to be larger on the left side to accommodate the strings). I had avoided making this assumption and was able to reduce the system to a fourth-order polynomial in one variable soon after.

I tell this story because it highlights the importance of teamwork in math: even if one person is clearly the smartest, problem solving is ultimately about creativity, and having more people thinking about a problem only means that 'aha!' moments/mistake catches will occur faster. This is a fantastic mindset to have if you, like myself, often spend time interacting with much smarter people.

As there may be a more elegant solution, I'd be happy to hear about the approach you took in the comments!

This is a great solution, but it's actually possible to solve the problem with only four variables! I split the vertical distance x into two variables, so the vertical height to the intersection point is h and the remaining height is x-h. The horizontal width along the bottom edge I called w, and along the top I called the horizontal width y.

ReplyDeleteNow, looking at the 10-foot and 12-foot web lines, we can create two large right triangles: w^2 + x^2 = 10^2, and x^2 + y^2 = 12^2. But this is only two equations and three variables, so we use the 5-foot horizontal line to create two similar triangles which will give us enough information to finish the problem.

There's a triangle with width 5 and height x-h, which is similar to the 10-foot-hypotenuse triangle, and one (upside-down) with width 5 and height h which is similar to the 12-foot-hypotenuse triangle. We only know the side lengths of these smaller triangles, but that's all we need to know: since the ratios of the sides are the same, we must have 5/w = (x-h)/x, and 5/y = x/h. From there you can combine the system of four equations to get a quartic equation in one of the variables, which solves to get the same result as in the post.