Fun questions :)

[yawho]

A fun question for your enjoyment.
The sides of a triangle are 13, 14 and 15. Find the exact radius of a circle which passes through the vertices of the triangle.

[moekamo]

radius = 65/8 units?

[kamil9876]

Given a list of real numbers greater than or equal to , prove that their product is greater than or equal to their sum.

e.g:

[Planck's constant]

51. Prove that the area of the annulus

Are we allowed to say that on this forum !

[rife168]

A circle is inscribed in a triangle whose sides are 40, 40 and 48 cm respectively. A smaller circle is tangent to the two equal sides of the triangle and to the first circle. Find the radius, in cm, of the smaller circle.

I've been working at this sporadically today, I think I can find the radius of the larger circle (not sure) but yeah I'm quite stumped.

[brightsky]

recall that the incentre is the intersection of the bisectors of the three angles.

now we have an isosceles triangle of dimensions 40 cm*40 cm*48 cm. draw this triangle so that its base has length 48 cm. by pythagoras' theorem, the height of the triangle is sqrt(40^2 - 24^2) = 32 cm. so it's area is 0.5*48*32 = 768 cm^2.

now draw in the angle bisectors. also draw in the three radii (r) of the circle perpendicular to the sides of triangle. the area of the triangle is hence given by:
0.5*40*r + 0.5*40*r + 0.5*48*r = 64r
this means that 64r = 768
r = 12

draw a tangent to the smaller circle parallel to the base of the triangle (this line is 24 cm above the base). we have now formed a smaller triangle similar to the big triangle. since this smaller triangle has height 8 cm, therefore, by the properties of similarity, the smaller triangle has dimensions 10 cm * 10 cm * 12 cm. we also known as a consequence that the area of the smaller triangle is 0.5*8*12 = 48 cm^2.

similar to what we did before, we let the radius of the smaller circle (which is in effect inscribed inside the smaller triangle) be x. this means that the area of the smaller triangle is given by:
0.5*12*x + 0.5*10*x + 0.5*10*x = 16x
16x = 48
x=3

hence the radius of the smaller circle is 3 cm.

[pi]

Forgive me if this has been answered

68. Simplify
(sauce: /0)

I love identities

[pi]

This is a complete non-mathematical stab in the dark (lol)

64. A one hectare field is in the shape of a right angled triangle. At the midpoint of each side is a post. over9000 the goat is tethered at the midpoint of the hypotenuse, and two sheep, TT and Mao, at the midpoints of the other two sides. Each rope is just exactly long enough to allow the animal to reach two adjacent vertices. One and only one animal can reach all points in the field. Which one is it and why?
(Image removed from quote.)

The goat for two reasons:
1) The sheep are pretty much in the same positions in relation to each other, and there is only one answer, so it can't be either of the sheep, hence the goat (over9000)
2) The sheep can't reach every point without passing over rope of the other sheep and/or the goat, and sheep (by their nature), just don't do that...


Dunno if that is right or if there is a mathematical answer haha

[rife168]

This is a complete non-mathematical stab in the dark (lol)

64. A one hectare field is in the shape of a right angled triangle. At the midpoint of each side is a post. over9000 the goat is tethered at the midpoint of the hypotenuse, and two sheep, TT and Mao, at the midpoints of the other two sides. Each rope is just exactly long enough to allow the animal to reach two adjacent vertices. One and only one animal can reach all points in the field. Which one is it and why?
(Image removed from quote.)

The goat for two reasons:
1) The sheep are pretty much in the same positions in relation to each other, and there is only one answer, so it can't be either of the sheep, hence the goat (over9000)
2) The sheep can't reach every point without passing over rope of the other sheep and/or the goat, and sheep (by their nature), just don't do that...


Dunno if that is right or if there is a mathematical answer haha

I'm fairly sure you are correct, the way I would do it would be to simply prove that the midpoint of the hypotenuse is equidistant to all vertices in a right angled triangle.
I remember doing this with vectors a while ago...

Anyway,
It is clear that the midpoint of the hypotenuse is equidistant from the two adjacent vertices, that is a condition of a midpoint.
Now we need to show that the right-angled vertex is the same distance from the midpoint of the hypotenuse.

This can be proven using the fact that the right angle of a right triangle always lies on the circumference of a circumcircle whose diameter is the hypotenuse. The centre of this circumcircle is the midpoint of the hypotenuse, and the right angle vertex is always the same distance from the centre of the circle (the radius). Hence all three vertices are equidistant from the midpoint of the hypotenuse of a right angled triangle

[Bhootnike]

I love identities

I love you vege, but i thought this was pretty funny :p 

I honestly don't remember enough of spesh to help you (ie. have no idea of identities anymore)

 

[pi]

I love identities

I love you vege, but i thought this was pretty funny :p 

I honestly don't remember enough of spesh to help you (ie. have no idea of identities anymore)

 

I had to google the identity for the 3rd last step, I knew it existed, just forgot the specifics!

Good call though haha

[pi]

btw, can someone do this one please (or give some hints), it's driving me insane!!!! I can't get very far into it (even after googling up the good ol' circle theorems again)

57. Find the missing angle
(Image removed from quote.)

[Mao]

btw, can someone do this one please (or give some hints), it's driving me insane!!!! I can't get very far into it (even after googling up the good ol' circle theorems again)

57. Find the missing angle
(Image removed from quote.)

Denote the "?" angle as x.
Looking at the shapes formed in the circle, there are two triangles. For now, let's take the upper triangle. The angle between the two radii is denoted a. Since the triangle is an isosceles, let the other angles be b. Similarly, 'd' denotes the angle between the two radii for the bottom triangle, and e is the other angle in the isosceles triangle.

We sum up the angles in the outer-most quadrilateral (the one including x):

a+b+x+d+e+50=360
x=310-a-b-d-e

However, from the triangles, we know that:
a+2b=180 ==> b=90-0.5a
d+2e=180 ==> e=90-0.5d

thus

x=310-a-(90-0.5a)-d-(90-0.5d)=130-0.5(a+d)

We also know that a+d+120+50=360 ==> a+d=190

Thus, x=130-95=35

[pi]

^+1, thanks a lot!

Didn't see this: a+b+x+d+e+50=360 (didn't see that quadrilateral) *sigh*

[TrueTears]

ok here's one: If i have a dice with 6 sides, in 6 rolls, how many unique numbers will be seen? In other words, what is the average number of unique numbers?

eg, if i roll a 1 1 1 1 1 1 that contains 1 unique number, if i roll a, 1 2 1 1 1 1 that contains 2 unique numbers, 1 2 3 4 5 6 ; 6 unique, etc

btw the die is 100% non biased.