Can Veeky Forums solve this?
The length of the chord [math]\overline{AB}[/math] is 20, and it intersects the chord [math]\overline{CD}[/math] at a right angle, splitting it in to segments [math]\overline{CX}[/math] and [math]\overline{XD}[/math] of lengths 3 and 12, respectively. What is the radius of the circle?
Also, challenge problem thread.
do your own goddamn homework
I'm in calc three, this is entirely recreational.
Another one:
The square [math] WXYZ [/math] is partitioned into four triangles of areas 2, 4, 6, and 8, in some order, by the point [math] P [/math]. What is the distance from [math] P [/math] to [math] O [/math], the center of the circle?
Forgot pic
(This ones pretty easy)
I got the square root of 120.25, but I'm on mobile so I don't wanna write latex
And one last one:
Let [math] a,b,c\in\{1,2,3,4,5,6,7,8,9\} [/math] be base-10 digits. Find all values of [math] a,~b,[/math] and [math] c [/math] such that [eqn] (aaaa)\cdot (aaaa) = cccccccc - bbbb [/eqn]
2. Easy shit
I got the same thing.
Here's an interesting problem.
I got 15.
>coordinate geometry
Absolute trash. If you are already using coordinates then nothing stops you from just bashing it. Shit problem.
13/4
[math] \sqrt{2.5} [/math]
a=3, c=1, b=2
It's okay to admit you couldn't figure it out
Wrong, wrong, and wrong.
How could the radius be less than half the length of the chord?
radius should be ((481)^(1/2))/2 or about 10.966
?
sqrt(4/5) ?
about 1,42
redid the calculations the radius is not 7.2, i got 16.7 from (276.84)^1/2
isn't this sqrt(2)? Got no pen and paper with me though so not sure.
From CX and XD, we have the power of X pp(X) as -36 and also that CD has length 15.
The intersecting chords theorem(Euclid III.35) allows us to deduce that AB is cut into lengths of 2 and 18. Say that AX has length 2.
X is 4.5 away from the midpoint of CD, and 8 away from AB midpoint. Thus, XO^2 = 64 + 20.25, where O is the centre of the circle.
From the definition of power of a point, we have r^2 = XD^2 - pp(X) = 120.25.
Here's a new puzzle.
Suppose we have two circles [math]c[/math] and [math]d[/math] where the image of [math]d[/math] under inversion wrt [math]c[/math] is itself. Show that [math]c[/math] is its own image under [math]d[/math]-inversion.
[math]\overline{AE}=\frac{1}{2}\overline{AB} = 10[/math]
[math]\overline{EF} = \frac{1}{2}\overline{CD} - \overline{CX} = 4.5[/math]
[math]r = \sqrt{10^2 + 4.5^2} = \sqrt{120.25}[/math]
could only find a lengthy solution:
>write Pythagoras equations of triangles AXD, AXC, CXB and DXB
>you got 4 equations with 4 unknowns
>solve
>you got for example length of [AX]
>find coordinates of the point E that's equally far from A, C and D in the the frame (X, (XD), (XA))
>calculate distance EC
>???
>profit
I did Geometry in high school and algebra and calculus in college. I don't remember ever doing shit like this.
That's a nice solution