Find the radius of a circle.

[anemone]

Points A, B, C, D are on a circle with radius R. \(\displaystyle |AC| = |AB| = 500\), while the ratio between \(\displaystyle |DC|\), \(\displaystyle |DA|\), \(\displaystyle |DB|\) is 1, 5, 7. Find R.

 

[Opalg]

Points A, B, C, D are on a circle with radius R. \(\displaystyle |AC| = |AB| = 500\), while the ratio between \(\displaystyle |DC|\), \(\displaystyle |DA|\), \(\displaystyle |DB|\) is 1, 5, 7. Find R.

Suppose that the distances DC, DA, DB are respectively 1, 5, 7, in some unknown units. Then AC and AB (which are known to be 500 in standard units) will be equal to $x$, say, in the unknown units.

The circle is the circumcircle of the triangles ABD and ACD. The circumradius of a triangle is always equal to one half of a side of the triangle divided by the sine of the opposite angle. Thus in the triangle ABD, $\sin(\angle BAD) = \dfrac7{2R}$; and in the triangle ACD, $\sin(\angle ADC) = \dfrac x{2R}.$

The cosine rule in triangle ABD gives $\cos(\angle BAD) = \dfrac{x^2-24}{10x}$; and in triangle ACD it gives $\cos(\angle ADC) = \dfrac{26-x^2}{10}.$

Now use $\sin^2 + \cos^2 = 1$ for those two angles to get the equations $$\frac{49}{4R^2} + \frac{(x^2-24)^2}{100x^2} = 1, \qquad \frac{x^2}{4R^2} + \frac{(26-x^2)^2}{100} = 1.$$ Multiply out the fractions to get $$25*49x^2 + R^2(x^2-24)^2 = 100R^2x^2, \qquad 25x^2 + R^2(26-x^2)^2 = 100R^2.$$ Subtract the second equation from the first to get $-96R^2x^2 + 25*48x^2 = 0$, which leads to $R = 5/\sqrt2$ (in the unknown units). Substitute that value for $R$ into one of the two equations, which then reduces to $(x^2-25)^2 = 0$. Thus $x=5$. But we know that $x=500$ in standard units, so we need to multiply by 100, giving $\boxed{R= 250\sqrt2}$ as the solution.

Edit. That solution is wrong: see below.

 

[Opalg]

Suppose that the distances DC, DA, DB are respectively 1, 5, 7, in some unknown units. Then AC and AB (which are known to be 500 in standard units) will be equal to $x$, say, in the unknown units.

The circle is the circumcircle of the triangles ABD and ACD. The circumradius of a triangle is always equal to one half of a side of the triangle divided by the sine of the opposite angle. Thus in the triangle ABD, $\sin(\angle BAD) = \dfrac7{2R}$; and in the triangle ACD, $\sin(\angle ADC) = \dfrac x{2R}.$

The cosine rule in triangle ABD gives $\cos(\angle BAD) = \dfrac{x^2-24}{10x}$; and in triangle ACD it gives $\cos(\angle ADC) = \dfrac{26-x^2}{10}.$

Now use $\sin^2 + \cos^2 = 1$ for those two angles to get the equations $$\frac{49}{4R^2} + \frac{(x^2-24)^2}{100x^2} = 1, \qquad \frac{x^2}{4R^2} + \frac{(26-x^2)^2}{100} = 1.$$ Multiply out the fractions to get $$25*49x^2 + R^2(x^2-24)^2 = 100R^2x^2, \qquad 25x^2 + R^2(26-x^2)^2 = 100R^2.$$ Subtract the second equation from the first to get $-96R^2x^2 + 25*48x^2 = 0$, which leads to $R = 5/\sqrt2$ (in the unknown units). Substitute that value for $R$ into one of the two equations, which then reduces to $(x^2-25)^2 = 0$. Thus $x=5$. But we know that $x=500$ in standard units, so we need to multiply by 100, giving $\boxed{R= 250\sqrt2}$ as the solution.

Did anyone notice that that solution cannot possibly be correct? If $x=5$ (in the unknown units) then AD = AC, which is geometrically impossible, because AD is obviously shorter than AC. What I did wrong was that the calculation for $R$ is correct, giving $R=5/\sqrt2$ (in the unknown units), but when you substitute that value to get an equation for $x$, the equation does not reduce to $(x^2-25)^2 = 0$, but to $(x^2-18)(x^2-32) = 0$. Thus $x=3\sqrt2$ or $4\sqrt2$. But $3\sqrt2<5$, which would mean that AC < AD. That can't happen if the points A, B, C, D occur in that order on the circle. Therefore $x=4\sqrt2$ in unknown units, and the scaling factor to convert unknown units to standard units is $500/(4\sqrt2).$ So the correct solution is $R = \dfrac5{\sqrt2}\,*\,\dfrac{500}{4\sqrt2} = \boxed{312.5}.$

 

[anemone]

Did anyone notice that that solution cannot possibly be correct? If $x=5$ (in the unknown units) then AD = AC, which is geometrically impossible, because AD is obviously shorter than AC. What I did wrong was that the calculation for $R$ is correct, giving $R=5/\sqrt2$ (in the unknown units), but when you substitute that value to get an equation for $x$, the equation does not reduce to $(x^2-25)^2 = 0$, but to $(x^2-18)(x^2-32) = 0$. Thus $x=3\sqrt2$ or $4\sqrt2$. But $3\sqrt2<5$, which would mean that AC < AD. That can't happen if the points A, B, C, D occur in that order on the circle. Therefore $x=4\sqrt2$ in unknown units, and the scaling factor to convert unknown units to standard units is $500/(4\sqrt2).$ So the correct solution is $R = \dfrac5{\sqrt2}\,*\,\dfrac{500}{4\sqrt2} = \boxed{312.5}.$

To be honest, I must confess to you that the value of R that I obtained was \(\displaystyle \frac{1250}{3}\)...

The time when I first saw your solution, I became very suspicious of my own result and I checked my working and found out I had made a premature assumption in my solution to which I didn't know how to make it right. I even thought to draw the circle out on a graph paper with the data that I obtained but I didn't do so at last. I told my boyfriend about this and he said I should ask for clarification if I was in doubt. But I thought to myself, 'Hey, Opalg is always right'...thus, it must be best to remain silent in this instance.

Having said this, here is my flaw solution and my mistake was by making false assumption that the center of the circle lies outside the cyclic quadrilateral ABDC and now, I even didn't sure if it was okay to label D in between B and C or should I make it lies between the points A and C...I must have thoroughly messed up the problem.



First, I let the the distances of DC, DA, and DB be a, 5a and 7a respectively with a is a positive real value.

By applying Cosine Rule to the triangle ACD, we have:
\(\displaystyle 500^2=a^2+(5a)^2-2(a)(5a) \cos (90^{\circ}-\alpha)\)

\(\displaystyle a^2(26-10 \sin \alpha) =500^2\) (1)

We do the same to the triangle ABD and obtained:
\(\displaystyle 500^2=(5a)^2+(7a)^2-2(5a)(7a) \cos (90^{\circ}-\alpha)\)

\(\displaystyle a^2(74-70 \sin \alpha) =500^2\) (2)

Solving the equations 1 and 2 for \(\displaystyle \sin \alpha\) yields

\(\displaystyle \sin \alpha = 0.8\)

Hence, \(\displaystyle a^2=\frac{125000}{9}\) or \(\displaystyle a=\frac{250\sqrt{2}}{3}\).

Last, by using the Cosine Rule again to the triangle OAB, we find that

\(\displaystyle R^2=R^2+500^2-2(R)(500)\cos \alpha\)

or

\(\displaystyle R=\frac{1250}{3}\).

 

[Opalg]

I even didn't sure if it was okay to label D in between B and C or should I make it lies between the points A and C...I must have thoroughly messed up the problem.

You certainly have not messed up on the problem. I think that the question of whether the points come in the order ABCD or ABDC as you go round the circle is absolutely critical. If they come in the order ABDC then that would correspond (in my solution) to the case where $x^2 = 18$, in which case I agree with your answer $R=1250/3$. I assumed that the order would be ABCD, in which case $x^2 = 32$ and $R = 1250/4$.