Geometry, no idea how to solve it

[trail jackal]

Imagine you have a right tringle with it‘s right angle-corner pointing up. Then it‘s left side is 3cm right side is 4 cm and the last side is 5cm. Now place a circle inside that tringle such that it touches the tringle twice. What is its’ radius?

[harsh barn]

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[white dirge]

Here's a picture. Suppose the lateral sides of the triangle are a and b (say, a < b).
If that's what you mean, what would x be?

[trail jackal]

Sry I formulated it incorrectly

Smth like that

[slow chasm]

You mean a semicircle.

Which is tangent to both legs.

[trail jackal]

Uh ye

[slow chasm]

Can you show us the original text of the problem?

[trail jackal]

It was a problem from my schools competition

I don‘t have the original q since the competition was online

[slow chasm]

Okay, so do you know what it means when a line is tangent to a circle?

[trail jackal]

Yes

It touches the circle only once

[slow chasm]

Do you know what else it means?

[trail jackal]

It has a right angle there?

[slow chasm]

A circle "has a right angle"?

A line "has a right angle"?

[trail jackal]

I mean the point where they intersect

But

That idea is false

So just ignore it

[slow chasm]

...a point "has a right angle"?

An angle is formed by two lines.

[trail jackal]

Ok

[slow chasm]

So if the first line is the tangent, what is the second line?

[white dirge]

Oh, I see.
Perhaps, this might help: draw a square like that and look at some similar triangles.

This setup seems very familiar for some reason...

Have I seen this before?

[slow chasm]

Spoiler.

[white dirge]

Oh yeah, I remember this problem now.
It's interesting, but quite long.

Well, not exactly like that, but very similar.

[slow chasm]

@trail jackal

[trail jackal]

The second line is tangent as well

[slow chasm]

...okay, the answer is the radius. A tangent to a circle is perpendicular to the radius of the circle at the point of tangency.

[slow chasm]

So then in your picture, draw the radii to which the sides of the triangle are perpendicular.

[trail jackal]

Is that correct

I think it is

[white dirge]

Yeah, I think that's correct.

So, in general, if we have lateral sides a and b, then r = 1/(1/a + 1/b). In other words, 1/r = 1/a + 1/b.

Pretty cool!

[trail jackal]

Yess😺

Damn if I just paid attention to similar tringles in this problems, that would be free 5 points

😔

[white dirge]

Well, don't worry! You got it, so that's good.

[trail jackal]

I got another problem from the competition. If you want to you can try it

Let S=2+4+6+…+100. Find out how many plus signs you atleast have to turn into negative signs such that S is negative.

[white dirge]

Oh, that's pretty interesting.

[white dirge]

Let's see. Let's denote two quantities:
x = 2 + 4 + ... + 2n
y = (2n + 2) + ... + 100
It's easy to see that x = n(n + 1), y = (50 - n)(51 + n). So:
x - y < 0
||n(n + 1) - (50 - n)(51 + n) < 0
2n^2 + 2n - 2550 < 0
n^2 + n - 1275 < 0
We need the smallest natural n, so let's just find the positive zero of that polynomial and round up.
n = ⌈(√(5101) - 1)/2⌉
Since 71^2 < 5101 < 72^2:
(71 - 1)/2 < (√(5101) - 1)/2 < (72 - 1)/2
35 < (√(5101) - 1)/2 < 35.5
Thus, n = 36.||

[trail jackal]

Ye it‘s the right way

It‘s n=35 but you‘re idea is correct

[white dirge]

Oh, really? Hm...

[trail jackal]

Yep

And then you just have to find out how many natural numbers are in the interval $(35;50]$

[vale charm]

Tony

[white dirge]

Well, 15.

[trail jackal]

Ye

[white dirge]

Ah, I think I see where I messed up.

I started counting from the wrong side 😅

[trail jackal]

Lol

+close

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