Problems With Everyday Objects: The Human Eye

For length of transverse common tangent(centres on opposite side) we have direct formula (Trevor's pro strats fails here :p)

Let distance between centre of 2 circles be 'd' And radius of constricted pupil 'r'

Let us solve for green line

We have

$\displaystyle L_{green} = \sqrt{d^{2} - (r_1 + r_2)^{2}}$

Here $r_1 = r_2 = 5r$

For red

$\displaystyle L_{red }= \sqrt{ d^{2} - (r_3 + r_4)^{2}}$

Here $r_3 = r, r_4 = 4r$

Solving (squaring)

$\displaystyle 2816 = d^{2} - 100r^{2}$

$\displaystyle 2891 = d^{2} - 25r^{2}$

Eliminating 'd' we get

$r = 1$

$4r = \boxed{\boxed{4}}$

I apologize in advance for the sheer amount of text as there is a lot of explaining to do. I will have a geometric diagram with these lines drawn tomorrow as I left my iPad at school and thus I can't make one at the moment. Hope you like my solution!

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Begin by drawing a line connecting the two circles centers, call the portion of this line that exists between the two blue circles d.

For now, label the green segment $G$ and the red segment $E$ .

What we are looking at here are two congruent and two similar triangles.

Working with the Green line

NOTE: all lines tangent to a circle form a 90 degree angle if a line is drawn from the point of tangency to the circle's center.

The line connecting the two circles centers will split the green line exactly in half. Infact, the point of intersection is also the midpoint between the two circles centers (aka, the mid point of d). This is because if you draw a line from both center's to their respective points of tangency, we have two right triangles with the same angles and the same bases, thus the triangles are congruent.

From this and the Pythagorean theorem, we get the equation $(b+\frac{d}{2})^2-b^2=\frac{G^2}{4}$ where b is the radius of the large blue circle.

Working with the Red Line

The red line is much harder. Call the radius of the small black circle r and the radius of the larger black circle R.

We already know that we will have two similar triangles due to the properties of tangent lines mentioned earlier.

However, this time, one base is four times the base of the other. Thus the length of the larger triangle's longer leg is 4 times the length of the smaller triangle's longer leg. The triangle with base $R$ 's longer leg will be of length $\frac{4E}{5}$ because this length + 1/4 this length must equal E.

The other leg will be of length $R$ and the hypotenuse will be $\frac{4}{5}$ the total distance between the circles centers, or $\frac{4(b+d+b)}{5}=\frac{8b+4d)}{5}$

Thus by Pythagorean theorem: $\frac{16E^2}{25}=(\frac{8b+4d)}{5})^2-R^2$

Combining the two

So far, we have established that $\frac{16E^2}{25}=(\frac{8b+4d}{5})^2+R^2$ and $\frac{G^2}{4}=b^2+(b+\frac{d}{2})^2$

In the problem, we are given that $4r=R$ and $5r=B$ . Substituting so we are left with d and r in the two equations above.

$\frac{16E^2}{25}=(\frac{40r+4d)}{5})^2-(4r)^2\Rightarrow r=\frac{1}{15} (\sqrt{d^2+3 E^2}-2d)$

$\frac{G^2}{4}=(5r+\frac{d}{2})^2-(5r)^2\Rightarrow r=\dfrac{G^2-d^2}{20 d}$

Now, we have to solve a relatively simple linear system, thus we must plug back in our original values of G and E.

$\frac{1}{15} (\sqrt{d^2+3 (7\sqrt{59})^2}-2d)=\dfrac{(16\sqrt{11})^2-d^2}{20 d}$

$4\sqrt{d^2+8673}-8d=8448-3d$

We could bash and solve this, or we can use a little bit of logic to cut a lot of work.

Dividing both sides by four, we see the LHS is $0 \pmod{4}$ , and since 8448 is (0 \pmod{4}) as well, d, must be divisible by four.

Now, the units digit for numbers divisible by four are 2,4,6,8, and 0. Upon squaring these digits and looking at the units digit, we get the digits 4,6, and 0.

Now, for $\sqrt{d^2+8673}$ to be an integer, ${d^2+8673}$ must end in a 0,1,4,9,6, or 5. We see that out of the set 0,4,6, only the number 6 when added to 8673 could possibly be a perfect square.

Thus we can check d=4,16,24,36,44. After checking these, we find that only d=44 yields an integral radical. (44 mm is 2 mm less than the average distance between eyes for a human)

Plugging our value of d back into our equation gives us

$r=\dfrac{(16\sqrt{11})^2-(44)^2}{20 (44)}\Rightarrow r=1$

$\therefore$ since the ratio of $r:R=1:4$ we have $R=4r=4(1)=\boxed{4}$

(Another fun fact: a dilated pupil tends to be between 3-4 mm in radius)

Let $r$ be the radius of the constricted pupil. Then the iris, (the blue circle), has radius $5r$ and the dilated pupil radius $4r$ . Also, let the distance between the centers of the two eyes be $L$ .

Now 'shift' the red segment downward, maintaining the same slope, so that its endpoint on the right coincides with the center of the eye on the right. This endpoint has thus moved a distance $4r$ , which means that, given the proportions of the eye on the left, the line segment will be tangent to this eye at $P$ . The triangle having vertices $P$ along with the two eye centers will then be a right triangle with legs length $5r$ and $7\sqrt{59}$ and hypotenuse length $L$ .

Thus $25r^{2} + 2891 = L^{2}$ , (A).

Next, 'shift' the green segment upward, maintaining the same slope, so that its endpoint on the right coincides with the center of the eye on the right. The endpoint $Q$ on the left will then be a distance of $10r$ from the center of the eye on the left, such that the triangle with vertices $Q$ and the two eye centers will be a right triangle with legs length $10r$ and $16\sqrt{11}$ and hypotenuse length $L$ .

Thus $100r^{2} + 2816 = L^{2}$ , (B).

Equating (A) and (B) gives us that

$100r^{2} - 25r^{2} = 2891 - 2816 \Longrightarrow 75r^{2} = 75 \Longrightarrow r = 1$ ,

as we must have $r \gt 0$ . Thus the radius of the dilated pupil is $4r = \boxed{4}$ .