NeoPets Riddles (Lenny Conundrums) and Answers Here

Uggg.... I hate waiting for the new LC. If I get in the top 250 this time I should get the gold trophy

i laready joined he lc 3x
but i cant get the av

new lenny
A local toy shop is making some large Kiko plushies. They took great care in designing the plushie's eyes:


The crosshairs are centered at the center of the eye. The black part of the eye is tangent to the edge of the eye, and the horizontal crosshair crosses 9 millimetres of the white part of the eye, and the vertical crosshair crosses 5 mm of the white part of the eye on both the top and bottom.

What is the area, in square millimetres, of the white part of the eye? Please round UP to the nearest whole number, and don't include any extra information or you run the risk of it being marked wrong. For example, if you calculate the answer to be 12.3 mm, simply submit 13 as your answer.

i got 333
i did 5+5+9 then divided by two to get the radius of the black circle
pi*r^2 for area

5+9=radius of white circle
pi*r^2

white - black = visible white area in square millimeters = 333

Not sure if i did that right though...it just seemed to make sense to me.

Wouldn't the radius of the white circle be the radius of the black circle plus 4.5? I'm probably wrong of course.

Yes, that's what I thought. I started by moving the black circle into the center of the white one, 4.5 units left...Am I headed in the right direction?

that does make sense...you move the black circle left 4.5 mm so that its centered in the white

You can ignore my first post then.

According to my ruler, 9mm splits the edge to center in half making the small circle have radius of 13.5 and larger with an 18...so 446 is what you get from that...

the radius of the black part would be 5.3125
since we know the it starts off with the vertical crosshair crosses 5 mm of the white part, then when moved 4.5 mm it becomes 4.5 itself

Now we would know the height of the arc to be .5 mm and the length of the chord would be 4.5

so to find the radius of the inner circle we would use the formula

r=(c²/8h)+(h/2)
r=(4.5²/8(.5))+(.5/2)
r=(20.25/4)+.25
r=5.0625+.25
r=5.3125

we can then figure out the area of both circles
black
a=pi(r²)
a=pi(5.3125²)
a=pi(28.22)
a=88.66

and for the radius of the white we would just add 4.5 to that of the black
a=pi(r²)
a=pi((5.3125+4.5)²)
a=pi(9.8125²)
a=pi(96.285)
a=302.49

then for the uncovered white part we simple subtract
white part=W-B
WP=302.49-88.66
WP=213.83

round up to 214

If I'm picturing this correctly, the radius of the white circle is 25mm, and the radius of the black circle is 20.5mm. Therefore, the white area is 643.24+ square mm, making the answer 644. It's an exercise in similar triangles.

Let the x- and y-axes be the crosshairs.
The equation for the white circle is x^2 + y^2 = 25^2.
The equation for the black circle is (x + 4.5)^2 + y^2 = 20.5^2.

Solve the latter equation for x=0 and you get y = +/-20 which leaves a white gap of 5mm (25-20) on the vertical crosshair.

That doesn't explain how I did it, but it allows you to verify the solution.

mark that formula won't work with this problem, and i'm guessing that you are not really sure of what the problem looks like so here is a pic of it

That is exactly as I had pictured it (mirror image, actually).

For your picture, the second equation would be:
(x - 4.5)^2 + y^2 = 20.5^2.

You can also use "power of the point." Consider the point inside the black circle where the crosshairs meet. On the horizontal crosshair, there are two segments. One has length R (radius of white circle), and the other has length R-9. On the vertical crosshair, the two segments each have length R-5. The "power of the point" tells us that:
R * (R-9) = (R-5) * (R-5)
Solve for R and you get 25mm as the radius of the white circle. Of course, the radius of the black circle is R - 9/2.

Explain why my answer is wrong.

well first they asked for the area of the white part of the eye not the whole white circle

and second i'm not really following how you came up with the radius of the white circle being 25 and that of the black being 20.5

ok i did screw something up with my formula
the chord should have been 9 not 4.5
so it should be
r=(c²/8h)+(h/2)
r=(9/8(.5))+(.5/2)
r=(81/4)+.25
r=20.25+.25
r=20.5

we can then figure out the area of both circles
black
a=pi(r²)
a=pi(20.5²)
a=pi(420.25)
a=1320.25

and for the radius of the white we would just add 4.5 to that of the black
a=pi(r²)
a=pi((20.5+4.5)²)
a=pi(25²)
a=pi(625)
a=1963.5

then for the uncovered white part we simple subtract
white part=W-B
WP=1963.5-1320.25
WP=643.25

so rounded up it would in fact be 644

i appologize mark and thank you for making me go back and correct my error

That's what I gave. Perhaps my wording was ambiguous.


That doesn't make it wrong.

I'll give the similar triangles solution. In your diagram, label the intersection of the vertical crosshair and the upper half of the circumference of the black circle A. Label the intersection of the x-axis and the left portion of the circumference of the black circle B. Label the intersection of the circumferences of the black and white circles C. Label the origin O.

ABC is a right triangle. OBA is a right triangle and is similar to ABC. (A=O=90, B=B).
Therefore, AB/OB = CB/AB
and AB^2 = OB * CB

OA = R-5 (given)
OB = R-9 (given)
OC = R (given)
CB = 2R-9 (addition)
AB^2 = OA^2 + OB^2 (Pythagoras)

OB * CB = OA^2 + OB^2
(R-9) * (2R-9) = (R-5)^2 + (R-9)^2
Solve for R.

I got that also
Yeah - I got that the radius of the big circle was 25mm and the small circle was 20.5 (doing my own calculations). So that would make the answer 644. I'm glad I remember geometry/algebra lol.

(the boring part of how I did it, probably same as you)

First:
Radius of Large Circle (RL) = segment of black (r1) + 9mm
Radius of Small Circle (RS) = segment of black (r1) + distance from center of large circle to center of small circle (r2).
RL also equals RS + r2. so...

RL = r1 + 9
RS = r1 + r2
RL = r1 + r2 + r2

Then by setting the two RL equations equal to one another, you get:
r1 + 9 = r1 + r2 + r2, which ends up as 9 = r2 + r2
therefore r2 = 4.5

Then, you can als have RL = segment from center of big circle to outside of black circle (T) + 5 mm
so then you can solve for T (or any other value you want) by making a triangle out of r2, T, and the RS.
solve for RS in terms of T: r1 + 9 = T + 5, so r1 = T - 4
r1 + r2 = RS, so (T - 4) + 4.5 = RS
using pythagorean theorem, you get...

(r2)^2 + T^2 = (T - 4 + 4.5)^2
4.5^2 + T^2 = T^2 + T + 0.25
20.25 = T + 0.25
T = 20
So since RL = T + 5, 20+5 = 25mm
And then solve for RS = 20.5

then pi r^2 with the white minus the black, = 643.24 = 644
I duno if that explanation was the same or clearer.

wow my brain just went back to HS
R*(R-9) = (R-5) * (R-5)
R²-9R = R²-10R+25
subtract R² from both sides
-9R = -10R+25
add 10R to both sides
R = 25

haven't done and equation like that in about 14 years

At least it's only been 14 years for you. :wink:

well I havent even learnt it! Im not even in high skool! let alone middle skool...

anyway is the answer 644?

I found the same solution, 643.2410955 mm
But I wonder, if you round up to the nearest whole number, isn't it 643?