Irrefutable proof that Veeky Forums is made up of brainlets who don't understand even basic, high-school geometry

Always ask yourself what information is being used, what isn't, and what will be necessary to incorporate

Clearly we can set up a system of equations to solve this problem using vectors
The question I suppose is if a more computationally convenient way exists

If you rotated the triangle around its centroid the trace the furthest point from it makes (that is, the most distant point on the r=5 semi circle) is the trace of the large circle. The large circle therefore has a radius of 5 plus the perpendicular length from the hypotenuse to the centroid.

Knowing the midpoint, you can then use the distance formula to find the length from the center of the circle to the middle of any of the sides of the triangle. Then simply add the radius of that particular side of the triangle.

I got sqrt(73/9) + 3 = 5.85

The radius of the circle is a rational number.

Just eyeballing it, my gut tells me 6.

Oh I see, I was assuming that the center of the triangle was the center of the circle.

no because the the midpoint of the triangle isn't necessarily the midpoint of the circle.

a,b,c,d,e,f,g,h,i,x are vectors

a,b,c are given and
i = (a+b)/2
h = (c+a)/2
g = (b+c)/2

We are to find d,e,f. The equation for a circle is (x-z)^2 = r^2
Or (x-e)^2 = (x-f)^2 = (x-d)^2

Also
(i-e)^2 = (a-i)^2
(g-f)^2 = (g-b)^2
(h-d)^2 = (a-h)^2

Also
(d-h)^2 + (g-f)^2 = (i-e)^2

That's 6 equations in 6 unknowns
have fun

Last equation follows from the three equations preceeding it. Someone would have to find another equation

>The equation for a circle is (x-z)^2 = r^2
>Or (x-e)^2 = (x-f)^2 = (x-d)^2

this is wrong because x isn't the midpoint of the circle, it's the midpoint of the triangle.

I think you have to find ex,dx and fx and then somehow use the angles between to determine the circle.

[math]\sqrt{50}[/math]

>Homework problem with 6 equations
OP what the fuck is wrong with your school? This shit is going to take like 15 minutes to find per problem like this.

This question comes from a 2 hour exam with 9 questions. It's from part A, the "easy" section. It's for high school students in grade 12.

>homework problem that takes 15 min is too long

You *really* don't belong here.

The short answer questions are not the "easy" section, they just don't require justification. A6 is usually the second hardest on the contest.

>Grade 12
That's very unusual. High school math competitions are generally open for all of high school and even middle school given how mathematical talent often demonstrates itself at a young age

For a high schooler it is, in America they have like 8 classes a day now.

I bet you could do something cool with the pythagorean theorem here by treating each circle as a circle inscribed into a square and shifted over s/2 units. I'll develop this idea and post results later

Tan, tan, tan rule you fucking brainlets

Like this

In case anyone hasn't figured it out, this is a specific case of the more general problem of apollonius. It involves using couple quadratics to find a circle tangent to three other circles. Radius is [math]\frac{144}{23}[/math]

the only way to solve this is to know the relation between triangles and circles, and there is one that's in my head right now
I can calculate it with knowledge of this relation if Ic an calculate the points d,e,f of course,
I think I can do it with area and some shortcut on the intersection arcs all adding to something
area of triangle
+ area of 3 semicircles
+ dcf, dae, fbe - but knowing these - there's probably some generalization here that means you can work it out without knowing the contact points.
then radius from area
but i can't recall the rule at this late hour
you don't know the tangents
you need to get the slopes between de, ef, df - and center from that
see incircles and excircles of a triangle
incenter here
en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Incenter

wrong
my answer is the only right one

...

Soo... whats the correct answer?

is this a thinly veiled gömböc thread

144/23 = 6.26

Area of Circle if r=6.26 is 123.11 (A)

On the other hand, area of triangle:
6*8 = 48

Area of Circle with Diameter 6: 28.26, semi-circle: 14.13
Area of Circle with Diameter 8: 50.24, semi-circle: 25.12
Area of Circle with Diameter 10: 78.5, semi-circle: 39.25

Adding all: 48 + 14.13 + 25.12 + 39.25 = 126.5 (B)

now A

Drew it out in CAD and measured the radius and circle center point

area of triangle is (6*8)/2, not 6*8

the triangle is 25 nigga

Yes, I goofed up. Thanks for pointing that out

Engineering brainlets will defend proof by cad

Mathcuck gonna devalue a computer's ability to magic

Which program?

Never claimed it was proof. It's just a solution.
Mastercam

12.53

Woops, I did diameter. Make that 6.265.

This is basic trig. Learn to into inscribed triangles. I doubt anyone on this board will have trouble understanding this. Sahgey.

if it is a rght triangle you take coordinates at the vertex and the solution is trivial.

I tutor high school geometry and it is the worst shit. I refuse to memorize all that gay shit.

>who don't understand even basic, high-school geometry.
I don't, I remember around 5% of geometry I've been taught and 95% of algebra. Is this common or maybe I'm just not a spatial thinker?

iti s right so the solution is 4+3=7

Here's how to do it if the points are known (which they are, as we have a 3,4,5 right triangle given):

[eqn]
I) ({X-{(A+B)\over{2}}})^2\ge({(B-A)\over{2}})^2\\
II) ({X-{(A+C)\over{2}}})^2\ge({(C-A)\over{2}})^2\\
III) ({X-{(B+C)\over{2}}})^2\ge({(C-B)\over{2}})^2\\
(M-X)^2=M^2-2MX+X^2\ge R^2\\
I+II+III:\\
X^2-2X{1\over3}(A+B+C)\ge0 \quad |+({1\over3}(A+B+C))^2\\
({(A+B+C)\over3}-X)^2\ge ({1\over3}(A+B+C))^2 = R^2\\
[/eqn]

Product of vectors denotes the dot product. This includes natural exponents.

>proof sci is a bunch of brainlets
>op doesnt give enough information
every time

[math]\frac{ \left(\frac{6+8+10}{2}\right)^2 }{ (6+8+10) -1}[/math]

:^)

see

asshole. beat me to it.

retards

op i checked and seem /sci are indeed brainlets

Look at all this gay math

Then look at this

This is a pretty cool problem

sqrt(73)/3+3

approximately 5.848

Geez, I got closer using a fucking ruler on my screen and notepad to do conversions.

I'm kind of amazed how wrong everyone who has posted is, its like you didn't even think about it at all.

The radius of the circle is higher than 6

amazing

Brilliant observation.
Pi is also around about 3.

Is there a proof to this tho?

I'm also wondering how he got that. Looks like black magic.

7.5

Can you explain your thought process? And perhaps background as you explain, that's

>that's
*thanks

I'm getting this.
If you draw a radius from the point where the big circle intersects a smaller circle, the radius will intersect the midpoint of the side corresponding to the smaller circle.
Use the pic from If you add (distance from the big circle's center to the midpoint of a side) + (radius of circle corresponding to the side), you should get the big circle's radius.

sqrt((x-3)^2 + y^2) + 3 = sqrt(x^2 + (y-4)^2) + 4 = sqrt((x-3)^2 + (y-4)^2) + 5 = R.

Where (x,y) is the center of the big circle and R is the radius.

(x-3)^2 + y^2 = (R-3)^2.
x^2 + (y-4)^2 = (R-4)^2.
(x-3)^2 + (y-4)^2 = (R-5)^2.

From these three equations you get 3x = 2y = R.
Using sqrt((x-3)^2 + y^2) = sqrt(x^2 + (y-4)^2) + 1, (the first two equations), substitute y = 3x/2, square both sides, isolate the square root, then square both sides again.
Solve the resulting linear equation for x then use R=3x.

I was just trolling, it's actually a miraculous coincidence it gives the correct answer like this. pure numerology and it only works for this specific Pythagorean triple lol

the problem is describes here
mathworld.wolfram.com/ApolloniusProblem.html

just plug the first 3 equations and corresponding circle centers in Mathematica and you'll get the correct answer. since this is a right triangle the answer is simpler.

radius is:
[math]\frac{2 a^3-a^2 b+2 a^2 c-a b^2-a b c+2 b^3+2 b^2 c}{8 a^2-9 a b+8 b^2} [/math]
where a,b,c are sides. the answer is [code] Solve[(x - a/2)^2 + (y - 0)^2 - (r - a/2)^2 == 0 && (x - 0)^2 + (y - b/2)^2 - (r - b/2)^2 == 0 && (x - a/2)^2 + (y - b/2)^2 - (r - Sqrt[a^2 + b^2]/2)^2 == 0 , {x, y, r}] [/code]

Hope you get an "A+" on your homework, friend.