Interesting math problem that I saw online

[anuttarasammyak]

@bob012345 You definitely get the correct numerical answer with your method, but I had no success in using your formula to get an answer for r in closed form in terms of a,b,c and d.

$$\sum_{n=1}^4 \sin^{-1} \frac{L_n}{d}=\pi$$
where d is diameter of circle and l_i are length of sides of quatrilateral. Using
sin(a+b)=sin a cos b + cos a sin b, and scaling $a_n=\frac{L_n}{d}$
We obtain three equations, as far as the quadrilateral contains center of the circle,
$$a_i\sqrt{1-a_j^2}+a_j\sqrt{1-a_i^2}=a_k\sqrt{1-a_l^2}+a_l\sqrt{1-a_k^2}$$
where {i,j,k,l}={1,2,3,4}. Otherwise
$$a_i\sqrt{1-a_j^2}+a_j\sqrt{1-a_i^2}=-a_k\sqrt{1-a_4^2}+a_4\sqrt{1-a_k^2}$$
where $a_1,a_2,a_3<a_4$ and {i,j,k}={1,2,3}.
I am not sure whether such transformation works well.

 

[Charles Link]

$$\sum_{n=1}^4 \sin^{-1} \frac{L_n}{d}=\pi$$
where d is diameter of circle and l_i are length of sides of quatrilateral. Using
sin(a+b)=sin a cos b + cos a sin b, and scaling $a_n=\frac{L_n}{d}$
We obtain three equations, as far as the quadrilateral contains center of the circle,
$$a_i\sqrt{1-a_j^2}+a_j\sqrt{1-a_i^2}=a_k\sqrt{1-a_l^2}+a_l\sqrt{1-a_k^2}$$
where {i,j,k,l}={1,2,3,4}
I am not sure whether such transformation works well.

This is @bob012345 's approach, and he got the algebra to work above, basically getting a closed form solution for $r$. In his latest post, (#27), he's only a step or two away from getting the general solution for $r$ in closed form.

 

[kuruman]

I. STRATEGY
1. Construct an arbitrary quadrilateral of fixed sides $(a,b,c,d)$ with hinged vertices.
2. Push (or pull) at opposing vertices as shown by the arrows in the drawing below until the sum of the angles $\theta +\varphi=\pi.$

Doing this makes the quadrilateral cyclic (inscribable in a circle.) In this drawing, $(a,b,c,d)=(2,3,4,5) ~\text{u}.$
3. Insert a diagonal rod $e$ and weld all vertices to make the quadrilateral rigid.
4. Find the area $A$ of the quadrilateral using Brahmagupta's formula.
5. Express this same area as the sum of the areas of two inscribed triangles in a circle of radius $R$.
6. Solve of $R$.

II. IMPLEMENTATION
Step 3

The figure on the right shows the rigid quadrilateral at step 3 above. We write the length squared of the diagonal $e$ using the law of cosines for each of the two triangles $$e^2=a^2+b^2-2ab\cos\theta$$ $$e^2=c^2+d^2-2cd\cos(\pi-\theta)$$This is a system of two equations and two unknowns which can be solved to give $$\cos\theta=\frac{(a^2+b^2)-(c^2+d^2)}{2(ab+cd)}$$ $$e=\left \{ (a^2+b^2)-\frac{ab}{(ab+cd)}\left[(a^2+b^2)-(c^2+d^2)\right] \right \}^{1/2}$$ For our choice of numbers, $e=4.411~\text{u}.$
Step 4
Brahmagupta's formula for the area of the cyclic quadrilateral is, $$A=\frac{1}{4}\sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}.$$ For our choice of numbers, $A=\sqrt{120}~\text{u}^2=10.95~\text{u}^2.$

Step 5
The area of a triangle circumscribed by a circle of radius $R$ is equal to the product of the three sides divided by $4R.$ Here we have $$\frac{abe}{4R}+\frac{cde}{4R}=A \implies R=\frac{e}{4A}(ab+cd).$$ Step 6
Using the above expressions for $e$ and $A$, $$R=\frac{ \left \{ (a^2+b^2)-\dfrac{ab}{(ab+cd)}\left[(a^2+b^2)-(c^2+d^2)\right] \right \}^{1/2}(ab+cd)}{\sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}}.$$ For our choice of numbers, $R=2.618~\text{u}.$

For $(2, 7, 11, 14)$, I get $R=7$ which makes the longest side a diameter. @Charles Link posted a variant of this here about a year ago.

(Edited for typos and embellishments)

 

[Charles Link]

@kuruman Excellent. :) You got $\cos{\theta}$ how I did, but once you have 3 points on the circle, (once you find $\cos{\theta}$ ), you can compute center $(h,k)$ and thereby you have $r$. Excellent solution you have in any case. :)

 

[anuttarasammyak]

sin(a+b)=sin a cos b + cos a sin b,

Instead let us use cos (a+b)=cos a cos b - sin s sin b for the case that quadrilateral contains conter of the circle
$$x_1x_2+x_3x_4=L_1L_2+L_3L_4:=A$$
$$x_2x_3+x_4x_1=L_2L_3+L_4L_1:=B$$
where
$$x_i:=\sqrt{d^2-L_i^2}$$
In the process of deletion of a x, we observe that x's are linear dependent.
$$\begin{pmatrix}
0 &B &L_2^2-L_4^2 &-A \\
B &0 &-A & L_1^2-L_3^2\\
L_4^2-L_2^2 &-A &0 &B \\
-A &L_3^2-L_1^2 &B &0 \\
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
x_3\\
x_4\\
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0\\
0\\
\end{pmatrix}
$$
By using this relation, we can delete three x's to leave one x in the equation and solve to get d. I have not done it yet.

[EDIT]
One more equation
$$x_1x_3+x_2x_4=L_1L_3+L_2L_4:=C$$
So I should rethink above said.

 

[kuruman]

@kuruman Excellent. :) You got $\cos{\theta}$ how I did, but once you have 3 points on the circle, (once you find $\cos{\theta}$ ), you can compute center $(h,k)$ and thereby you have $r$. Excellent solution you have in any case. :)

Thanks. That's the way I saw an easy-to-explain path to the answer. I shied away from spending time to cast the answer in a symmetric form under side swap. This is done in the Wikipedia article where I got the screenshot below.

 

[bob012345]

Starting from the generalized form of my original formula

$\arcsin(\frac{ax}{2})+ \arcsin(\frac{bx}{2})+ \arcsin(\frac{cx}{2})+ \arcsin(\frac{dx}{2})=\pi.$

we get;
$\sqrt{\left(1-\frac{a^2x^2}{4}\right) \left(1-\frac{b^2x^2}{4}\right)}\\ \quad+ \sqrt{\left(1-\frac{c^2x^2}{4}\right) \left(1-\frac{d^2x^2}{4}\right)} =\frac{(ab+cd)x^2}{4}$
Define. $u=\frac{x^2}{4}$

For convenience, let

$F=\sqrt{(1-a^2u)(1-b^2u)}$
and

$G=\sqrt{(1-c^2u)(1-d^2u)}$

Our equation is now simply

$F+G=(ab+cd)u$
Squaring gives

$F^2+2FG+G^2=(ab+cd)^2u^2$

Expand the individual terms:

$F^2=1-(a^2+b^2)u+a^2b^2u^2$ and $G^2=1-(c^2+d^2)u+c^2d^2u^2$

Substituting and rearranging,

$2FG=-2+(a^2+b^2+c^2+d^2)u+2abcd\,u^2$

Therefore,

$FG=-1+\frac{a^2+b^2+c^2+d^2}{2}u+abcd\,u^2$

Squaring the last equation gives

$F^2G^2= \left[-1+\frac{a^2+b^2+c^2+d^2}{2}u+abcd\,u^2\right]^2$

Substitute the expressions for $F^2$and$G^2$:

$(1-a^2u)(1-b^2u)(1-c^2u)(1-d^2u)\\ =\left[-1+\frac{a^2+b^2+c^2+d^2}{2}u+abcd\,u^2\right]^2$

This has no square roots but is a bit of a challenge to expand and simplify!

1. Left-Hand Side (LHS)​

$\text{LHS} = (1 - a^2u)(1 - b^2u)(1 - c^2u)(1 - d^2u)$

Expanded out term-by-term in powers of $u$
$u^0$ 1
$u^1$:$- (a^2 + b^2 + c^2 + d^2)$
$u^2$: $+ (a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2)$
$u^3$:$- (a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2)$
$u^4$:$+ a^2b^2c^2d^2 u^4$

2. Right-Hand Side (RHS)​

$\text{RHS} = \left[-1 + \frac{a^2+b^2+c^2+d^2}{2}u + abcd,u^2\right]^2$

Expanded out term-by-term in powers of $u$]:
$u^0$:$= 1$
$u^1$:$- (a^2 + b^2 + c^2 + d^2)$
$u^2$:$+ \left[\frac{(a^2+b^2+c^2+d^2)^2}{4} - 2abcd\right]$
$u^3$:$+ abcd(a^2 + b^2 + c^2 + d^2)$
$u^4$:$+ a^2b^2c^2d^2$

The zeroth, first and fourth powers of $u$ cancel out. Then $u$ will simply be (haha) the difference of the coefficients of the second power divided by the difference of the coefficients of the third power of $u$.

$u =\Large \frac{2abcd + \frac{1}{2}\left(a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2\right) - \frac{1}{4}\left(a^4 + b^4 + c^4 + d^4\right)}{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2 + abcd\left(a^2 + b^2 + c^2 + d^2\right)}$

The numerator factors as;
$N=\frac{-1}{4}(a+b+c−d)(a+b−c+d)(a−b+c+d)(a−b−c−d)$

or using the semiperimeter;
$4(s-a)(s-b)(s-c)(s-d)$

The denominator factors as;
$D=(ab+cd)(ac+bd)(ad+bc)$

The complete form is;

$u=\Large\frac{4(s-a)(s-b)(s-c)(s-d)}{(ab+cd)(ac+bd)(ad+bc)}$

Now, $u=\frac{x^2}{4}=\frac{1}{4r^2}$

Thus $r=\sqrt{\frac{1}{4u}}$

This gives us $r=\Large\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$

I think I prefer @Charles Link’s method!

 

[Charles Link]

Instead let us use cos (a+b)=cos a cos b - sin s sin b for the case that quadrilateral contains conter of the circle

This is from post 35 above. I don't know why $\sin(a+b)$ doesn't seem to work on @bob012345 's formula. That's what I tried around post 28 and I found it to be very messy. In any case super job @bob012345 and also @kuruman and also @anuttarasammyak for some very healthy responses. :)

 

[anuttarasammyak]

continued form #31

We get three equations

$$\begin{pmatrix}

L_2 & L_1 & -L_4 & -L_3 \\

L_3 & -L_4 & L_1 & -L_2 \\

L_4 & -L_3 & -L_2 & L_1 \\

\end{pmatrix}

\begin{pmatrix}

x_1 \\

x_2 \\

x_3\\

x_4\\

\end{pmatrix}

=

\begin{pmatrix}

0 \\

0\\

0\\

\end{pmatrix}$$
where
$$x_i:=\sqrt{d^2-L_i^2}$$

By adding one row on the matrix we can make it symmetric 4X4

$$\begin{pmatrix}

L_1 & L_2 & L_3 & L_4 \\

L_2 & L_1 & -L_4 & -L_3 \\

L_3 & -L_4 & L_1 & -L_2 \\

L_4 & -L_3 & -L_2 & L_1 \\

\end{pmatrix}

\begin{pmatrix}

x_1 \\

x_2 \\

x_3\\

x_4\\

\end{pmatrix}

=

\begin{pmatrix}

4A \\
0 \\

0\\

0\\

\end{pmatrix}$$
where A is area of quadrilateral.

$$\begin{pmatrix}

x_1 \\

x_2 \\

x_3\\

x_4\\

\end{pmatrix}


=\frac{1}{L_1^2-L_2^2-L_3^4-L_4^2}


\begin{pmatrix}


L_1 & -L_2 & -L_3 & -L_4 \\

-L_2 & L_1 & L_4 & L_3 \\

-L_3 & L_4 & L_1 & L_2 \\

-L_4 & L_3 & L_2 & L_1 \\

\end{pmatrix}
\begin{pmatrix}

4A \\
0 \\

0\\

0\\

\end{pmatrix}$$

some x are minus. There is something wrong.

 

[bob012345]

This is from post 35 above. I don't know why $\sin(a+b)$ doesn't seem to work on @bob012345 's formula. That's what I tried around post 28 and I found it to be very messy. In any case super job @bob012345 and also @kuruman and also @anuttarasammyak for some very healthy responses. :)

Since my original equation uses arcsines, and because of the differences in sine and cosine addition formulas, you get twice as many radicals to get rid of if you take the sine of both sides than if you take cosines. Also, this is if you arrange the equation as
$A+B=\pi-(C+D)$ first. In both cases that allows some terms to cancel out. Try taking $\sin(A+B+C+D)=\sin(\pi)=0$ and you may end up with an eight radical beast!

 

[Charles Link]

@anuttarasammyak I don't know that your method (referring to post 39) is valid. The $x's$ are not variables with any freedom. You only need one expression, not 3 of them to solve for $d$.

Edit: Meanwhile @bob012345 worked through all the difficult algebra on this one successfully, and he only needed one expression, and not the other two permutations.

 

[anuttarasammyak]

Thanks for the teachings. I might have failed for null vevtors in #39.

Putting d＝0 in the general solution I got familiar formula for triangle. It is a good check for me.

I would likt to confirm that the general solution does not depends on whether quadrilateral contains O which is center of the circle or not.

In case there is no circumcircle, how the general solution responds ？

 

[kuruman]

I would likt to confirm that the general solution does not depends on whether quadrilateral contains O which is center of the circle or not.

View attachment 374833

Confirmed. The solution that I provided in post #33 is applicable to either of the inscribed quadrilaterals that you show in your first figure. The position of the circle's center is irrelevant as far as the derivation is concerned.

In case there is no circumcircle, how the general solution responds ？

The statement of the problem in post #1 begins with "You have a 4 sided polygon inscribed in a circle" and asks for the radius of this circle. The general solution addresses and answers that specific question. If you have a quadrilateral that cannot be inscribed in a circle, then the circle does not exist. What exactly could we be asked to find then?

 

[anuttarasammyak]

Confirmed. The solution that I provided in post #33 is applicable to either of the inscribed quadrilaterals that you show in your first figure. The position of the circle's center is irrelevant as far as the derivation is concerned.

The general solution addresses and answers that specific question.

Thanks, I got it. I noticed that we can get unike circumcircle radius for any given a,b,c,d >0 by transforming angles of quadrilateral.

II. IMPLEMENTATION
Step 3

The figure on the right shows the rigid quadrilateral at step 3 above. We write the length squared of the diagonal e using the law of cosines for each of the two triangles e2=a2+b2−2abcos⁡θ e2=c2+d2−2cdcos⁡(π−θ)This is a system of two equations and two unknowns which can be solved to give cos⁡θ=(a2+b2)−(c2+d2)2(ab+cd) e={(a2+b2)−ab(ab+cd)[(a2+b2)−(c2+d2)]}1/2 For our choice of numbers, e=4.411 u.

$$e=\sqrt{\frac{ab(c^2+d^2)+cd(a^2+b^2)}{ab+cd}}$$

Circumradius of triangle sides of which are a,b,e is,
$$r=\frac{abe}{\sqrt{(a+b+e)(b+e-a)(e+a-b)(a+b-e)}}$$ $$=\frac{abe}{\sqrt{[(a+b)^2-e^2][e^2-(a-b)^2]}}$$
$$=\sqrt{\frac{(ab+cd)[ab(c^2+d^2)+cd(a^2+b^2)]}{4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2}}$$
$$=\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{8abcd+2(a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2)-(a^4+b^4+c^4+d^4)}}$$
$$=\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(b+c+d-a)(c+d+a-b)(d+a+b-c)(a+b+c-d)}}$$

Another approach
$$e=\sqrt{\frac{(ac+bd)(ad+bc)}{ab+cd}}$$
Familiar circumcircled triangle relation
$$r=\frac{abe}{4A_{abe}}=\frac{cde}{4A_{cde}}$$
where A_{abe} is area of triangle abe and so on.
$$r(4A_{abe}+4A_{cde})=(ab+cd)e=\sqrt{(ab+cd)(ac+bd)(ad+bc)}$$
$$r=\frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4A_{abcd}}$$
where A_{abcd} is area of circumcircled quadlilateral.　After some calculation we can prove that
$$4A_{abcd}=\sqrt{(b+c+d-a)(c+d+a-b)(d+a+b-c)(a+b+c-d)}$$

 

[Charles Link]

Thanks for the teachings. I might have failed for null vevtors in #39.

Putting d＝0 in the general solution I got familiar formula for triangle. It is a good check for me.

I would likt to confirm that the general solution does not depends on whether quadrilateral contains O which is center of the circle or not.

View attachment 374833

In case there is no circumcircle, how the general solution responds ？
View attachment 374834

By $d$ you are referring to the 4th side, and not the diameter of the circle. That took me a second look to figure out what you were referring to.

 

[Charles Link]

I would likt to confirm that the general solution does not depends on whether quadrilateral contains O which is center of the circle or not.

Notice when $a+b+c-d$ is small, when the 4 sided polygon is very flat, that $r$ is very large. The center of the circle is then very far from the polygon and is far outside of it.

(This is just one example. The 3 other terms in the denominator all have a similar effect. Here I assumed $d$ is the longest side.)

 

[anuttarasammyak]

After some calculation we can prove that

$$2A_{abcd}=(ab+cd)\sin\theta = (ab+cd)\sqrt{1-(\frac{a^2+b^2-c^2-d^2}{2(ab+cd)})^2}$$
$$=\frac{1}{2}\sqrt{4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2}$$
$$4A_{abcd}=\sqrt{(b+c+d-a)(c+d+a-b)(d+a+b-c)(a+b+c-d)}$$

 

[kuruman]

$$2A_{abcd}=(ab+cd)\sin\theta = (ab+cd)\sqrt{1-(\frac{a^2+b^2-c^2-d^2}{2(ab+cd)})^2}$$
$$=\frac{1}{2}\sqrt{4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2}$$
$$4A_{abcd}=\sqrt{(b+d+c-a)(c+d+a-b)(d+a+b-c)(a+b+c-d)}$$

Yes, that is Brahmagupta's formula quoted in post #33.

 

[bob012345]

If anyone is interested, here is a special case that is much easier to solve. Let two points be opposite on the circle. The other two can be arbitrary. Of course it’s not as easy to solve as a square or rectangle. The side lengths can all be different.

 

[kuruman]

If anyone is interested, here is a special case that is much easier to solve. Let two points be opposite on the circle. The other two can be arbitrary. Of course it’s not as easy to solve as a square or rectangle. The side lengths can all be different.

Can you be more specific? What is given and what are we looking for?

For example, if the two points that are "opposite on the circle" are diametrically opposed and, as in this thread, we are looking for the radius of the circle, then the answer is "the halfway distance between the two given points and who cares about the other two?"

 

[bob012345]

Can you be more specific? What is given and what are we looking for?

For example, if the two points that are "opposite on the circle" are diametrically opposed and, as in this thread, we are looking for the radius of the circle, then the answer is "the halfway distance between the two given points and who cares about the other two?"

Here is one example out of an infinite number. Points P2 and P4 are opposite on the circle. We only know the lengths of chords a,b,c and d. We don’t know the location of the points or the radius of the circle. Determine the radius in terms of the chord lengths.

 

[bob012345]

It doesn’t matter but I really should have drawn it with a,b on one side of the diameter and c,d on the other. If anyone works it feel free to change it around. The problem isn’t hard but I found a little surprise.

 

[kuruman]

Well, since the line joining P₂ and P₄ is a diameter, the triangles on either side of it are right triangles no matter where you put P₁ and P₃. Referring to the diagram in post #51, the Pythagorean theorem says, $$b^2+c^2=a^2+d^2=4R^2\implies R=\frac{1}{2}\sqrt{b^2+c^2}= \frac{1}{2}\sqrt{a^2+d^2}. $$ I agree, the problem isn't hard. What is the surprise?

 

[bob012345]

Well, since the line joining P₂ and P₄ is a diameter, the triangles on either side of it are right triangles no matter where you put P₁ and P₃. Referring to the diagram in post #51, the Pythagorean theorem says, $$b^2+c^2=a^2+d^2=4R^2\implies R=\frac{1}{2}\sqrt{b^2+c^2}= \frac{1}{2}\sqrt{a^2+d^2}. $$ I agree, the problem isn't hard. What is the surprise?

Ok, I didn’t realize they were always right triangles so I thought it was interesting that the sum of the squares of the chords on either side of the diameter were equal. In other words, I didn’t use that fact to get the radius.

 

[bob012345]

At least I didn’t propose the case where one of the chords goes through the origin!

 

[anuttarasammyak]

∑n=14sin−1⁡Lnd=π

In generalc case of m polygon
$$\sum_{i=1}^m \sin^{-1} \frac{L_i}{d}=\pi$$
seems to hold with choice of parameter d for any ${L_1,L_2,..,L_m}$. So there seems to exist a circumcircle.

 

[kuruman]

In generalc case of m polygon
$$\sum_{i=1}^m \sin^{-1} \frac{L_i}{d}=\pi$$
seems to hold with choice of parameter d for any ${L_1,L_2,..,L_m}$. So there seems to exist a circumcircle.

But there is no circumcircle in the case that you yourself proposed in post #42:

If you draw the 4 perpendicular bisectors they will not all intersect at the same point.

Also, what is this parameter $d$ and how is it related to the number of sides $m$?

 

[anuttarasammyak]

In generalc case of m polygon
∑i=1msin−1⁡Lid=π
seems to hold with choice of parameter d for any L1,L2,..,Lm. So there seems to exist a circumcircle.

Where d is diameter of the circle re:#31.


Correction: Any side must be shorter than sum of all the other sides. Heron and Brahmagupta area must be real number.

 

[kuruman]

OK, but your equation $$\sum_{i=1}^m \sin^{-1} \frac{L_i}{d}=\pi$$ is derived by assuming that the polygon is already inscribed in a circle. Therefore the statement

So there seems to exist a circumcircle.

cannot be a conclusion because it is an assumption in your argument.

The equation is a necessary and sufficient condition for the polygon to be inscribable in a circle of diameter $d$: If the equation holds, the polygon can be inscribed in a circle; if the polygon can be inscribed in a circle, the equation holds.

In the general case of a polygon, the equation does not necessarily hold. As we have seen in the case of your second polygon in post #42, you cannot find a circle that can pass through all four vertices of the quadrilateral.

 

[anuttarasammyak]

There are m rods length of each does not surpass sum of the others. With these rods we can make m polygons which has a circumcircle. It seems to hold.