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redacted123

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« on: November 04, 2009, 03:15:27 pm »
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« Last Edit: December 25, 2015, 05:31:38 pm by Stany »
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Jim Groovester

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Re: Math problem
« Reply #1 on: November 04, 2009, 03:36:14 pm »
It's not a variable, so it won't change. It's a fixed number.

Can you tell which segment is the hypotenuse? It would make this easier.

All right triangles must satisfy the pythagorean theorem. I'm guessing that AB is the hypotenuse, so:

(AB)2 = (AC)2 + (BC)2

To find the lengths of these segments, we must use the distance formula, which is basically just the pythagorean theorem again. Using substitution because writing square root signs is difficult:

((7 + 1)2 + (2 + 2)2) = ((k + 1)2 + (4 + 2)2) + ((7 - k)2 + (4 - 2)2)

(The square from the theorem and the square root from the distance formula cancel each other out, in case you were wondering.)

80 = k2 + 2k + 1 + 36 + 49 - 14k + k2 + 4

We then solve for k:

2k2 - 12k + 10 = 0
2(k - 5)(k - 1) = 0

This means that k can be either 5 or 1. And both work, both produce right triangles.
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Quote from: Jim Groovester on April 16, 2021, 04:58:54 pm
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redacted123

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« Reply #2 on: November 04, 2009, 03:44:30 pm »
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« Last Edit: December 25, 2015, 05:32:02 pm by Stany »
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Vector

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Re: Math problem
« Reply #3 on: November 04, 2009, 03:51:43 pm »
But here's the thing--we can't assume that AB is the hypotenuse.  We've got a bunch of potential vertices running around.

My guess is that, rather than "constant," the teacher meant "natural number"--because k had better not be a variable, if this is some particular right triangle we're looking for.

So, I've got a few numbers lying around, but as-is the question makes very little sense.  I also get a right triangle for k = 6 and AC as hypotenuse.
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Virex

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Re: Math problem
« Reply #4 on: November 04, 2009, 04:53:59 pm »
I've been doodling a bit (using digital graph paper: http://www.geogebra.org/cms/index.php?lang=en) and it seems there are 4 options. First of all you can assume that corner C is straight, which gives you 2 points, each of which is an intersection between a circle with AB as diameter and the line y = 4. These are already given by Jim.

Then there are two more sollutions, one for angle A is straight and one more for angle B is straight. These are given by :

(BC)2 = (AC)2 + (AB)2
and
(AC)2 = (BC)2 + (AC)2
Both of which by now should be trivial. They should also both only yield a single awnser.

(note, the anwser K = 6 is for C as a straight angle, so you're given that one as well)
« Last Edit: November 04, 2009, 05:14:35 pm by Virex »
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redacted123

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« Reply #5 on: November 04, 2009, 04:57:49 pm »
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« Last Edit: December 25, 2015, 05:32:26 pm by Stany »
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Muz

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Re: Math problem
« Reply #6 on: November 06, 2009, 11:44:38 am »
the side opposite to the right angle is the hypotenuse. Or whichever side is the longest :P
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redacted123

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« Reply #7 on: November 06, 2009, 04:29:12 pm »
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« Last Edit: December 25, 2015, 05:32:43 pm by Stany »
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Virex

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Re: Math problem
« Reply #8 on: November 06, 2009, 07:09:45 pm »
Quote from: Stany on November 06, 2009, 04:29:12 pm
Quote from: Muz on November 06, 2009, 11:44:38 am
the side opposite to the right angle is the hypotenuse. Or whichever side is the longest :P
Obviously, but because of the vagaries of the C co-ordinate, it wasn't immediately clear.

You mean that it's clear now? I still see three options ;)
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AtomicPaperclip

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Re: Math problem
« Reply #9 on: November 09, 2009, 10:52:34 pm »
Do you really need to do it so algebraically?

I drew the two known points, and drew a line at y=4. From there you can make a right triangle from each angle.

So just from just making perpendicular lines you can figure out that you can make a triangle out of the points (-2.5, 4) and (6,4), right?
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