Math Is Fun Forum

ET ag

Member

Registered: 2017-02-22

Posts: 10

Hard Proof

Hi,

I need a proof for the following problem before the midnight of Mar 3. I don't need a complete answer, though it would be helpful, since I need to write this in my own words. I need help PROVING, not knowing. So try to make it rigorous so that a person that has enough math knowledge to understand the problem and knows coordinate geometry can understand. However, don't "imply" anything or write it so that only a person with a PhD in Mattematics can understand.

Here's the question to prove: Let $A$, $B$, and $C$ be three points on the curve $xy = 1$ (which is a hyperbola). Prove that the orthocenter of triangle $ABC$ also lies on the curve $xy = 1$.

My feedback on this question: I know what an orthocenter is: where the three altitudes of a triangle meet, and it makes sense that it lies on a hyperbola. (Since I tried this on a piece of paper). But I can't just say, "If you try it, you'll see that this is possible." I must PROVE that this is possible and that is what I need help with. I explained above the deadline and the requirements, guidelines, and suggestions.

Thanks,
ET ag

P.S. I will check this thread every so often, so I should take too long to reply.

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If a second was a minute, and a minute was an hour, how many hours would be in one day? (There are multiple answers and they are debatable)

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Hard Proof

Hint:Take O(h,k) as orthocenter.Let A(x1,y1),B(x2,y2) C(x3,y3)
multiply the slopes of OA and BC and set it to -1.One more equation for OB.Solve simultaneously. you will get h=y1y2y3 and k=x1x2x3. Product hk=1 and we are done.
The work you have to carry through.

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Hard Proof

hi ET ag

My method is similar.  I called the points (a,1/a); (b,1/b); and (c,1/c) and worked out the equation of the line perpendicular to AB through C.  Once you have the form for that line, you can write down a similar equation for the line perpendicular to BC through A.

Find where they intersect (most of the algebra cancels out) and show the resulting point is on the hyperbola.  [This also shows that the third altitude goes through the same point.]

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

ET ag

Member

Registered: 2017-02-22

Posts: 10

Re: Hard Proof

Hi,

multiply the slopes of OA and BC and set it to -1

What exactly does this mean?

Also, from what the rest of thickhead's method is, how exactly should I find the coordinates of (h, k). Sorry, I just didn't really understand what you wrote.

Lastly,

  I called the points (a,1/a); (b,1/b); and (c,1/c) and worked out the equation of the line perpendicular to AB through C.  Once you have the form for that line, you can write down a similar equation for the line perpendicular to BC through A.

How did you work the equation for the line perpendicular to Ab through C?

Thanks,
ET ag

P.S. I only need a specification, since I see the method that you guys used, which was quite smart.

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If a second was a minute, and a minute was an hour, how many hours would be in one day? (There are multiple answers and they are debatable)

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Hard Proof

Last edited by thickhead (2017-02-24 04:24:45)

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Hard Proof

How did you work the equation for the line perpendicular to Ab through C?

If two lines are perpendicular then the product of their gradients is always -1  **

So if you work out the gradient of AB , let's say it is m, then the line perpendicular to AB has gradient -1/m

So the equation will be y = -x/m + k and you can work out k as you want the line to go through C.

It's worth simplifying as you go because the final result is fairly simple.

Bob

** If you have not met this before and want a proof, please post back.

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

ET ag

Member

Registered: 2017-02-22

Posts: 10

Re: Hard Proof

Ok. Thanks, I don't need a proof for **. I am in a little busy right now, so I will post my question later.

Thanks,
ET ag

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If a second was a minute, and a minute was an hour, how many hours would be in one day? (There are multiple answers and they are debatable)