Math Is Fun Forum

Enshrouded_

Member

Registered: 2015-07-31

Posts: 47

Geometry Problem

Point G is the midpoint of median

of

. Point

is the midpoint of

, and point

is the intersection of

and

. Find the area of

if

.

I've tried at this problem for a while, and I can't get it. I know many midpoints and that point T seems to be important. Thanks!

Last edited by Enshrouded_ (2015-08-01 00:45:20)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Geometry Problem

Hi;

Please use latex for this forum:

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Geometry Problem

hi Enshrouded_

Welcome to the forum.

I understood the diagram without any need for overlining etc.

In a triangle that is split in two by a line to the midpoint of the base,  the area of the two parts will be equal.

So if XYM = 150, then XYM = 75, and MHX = 75/2

What follows is wrong.  Solution may appear in a later post with any luck.

Triangle MTG is mathematically similar to MHX with a length scale factor of 1/2, so the area scale factor will be I/2 x 1/2 = 1/4.

Bob

---

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

Enshrouded_

Member

Registered: 2015-07-31

Posts: 47

Re: Geometry Problem

You mean XYZ = 150. My first try was actually 9 3/8, which turned out to be wrong. How can you tell that MTG is 1/2 of MHX?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Geometry Problem

Yes I did mean XYZ.

And now I look again, I have fiddled that last step.  I'll have another try now.

Bob

---

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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Geometry Problem

I see where I'm going wrong.  I read that as T is a midpoint too.  Now I've constructed it properly in Sketchpad, I can see that is not true.  I'm going to edit my earlier post to admit it is wrong, in case anyone is misled.  Then I'll try this again with the right diagram.

Back after I've pruned some apple trees.

Bob

---

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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Geometry Problem

I'm now getting 12.5 But I've got a difficulty proving it. 

Bob

---

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

Enshrouded_

Member

Registered: 2015-07-31

Posts: 47

Re: Geometry Problem

I see now. In XYM, YG and MH are medians, so they divide the triangle into 6 smaller and equal area triangles. So MTG = 1/6 XYM = 25/2. Thanks bob!

Enshrouded_

Member

Registered: 2015-07-31

Posts: 47

Re: Geometry Problem

Last question for a little bit:
In triangle

,

. The length of median

is 5. Let

be the largest possible value of

, and let

be the smallest possible value. Find

.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Geometry Problem

I finally sorted out my error and did it by vector geometry.  You were right that T is the key. I calculated That YT : TG = 2 : 1 and the result follows from that.  Well done for getting it by your own method.

Your last question has me puzzled.  I played around with the cosine rule for a bit but ended up with an expression for AB^2 - AC^2 so that didn't help.  I thought I'd use Sketchpad again to get a feel for when the max and min occur and found that AB^2 + AC^2 is a constant for all positions of A.

I made the line BC and found the midpoint, D.  I constructed a circle centre D and radius 5 with A on the circle.  Then I calculated AB^2 + AC^2.  I got 81.94 for all A on the circle.

If that's correct then max = min and M-n = 0.  Somehow I don't think that is correct. 

Bob

---

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

Enshrouded_

Member

Registered: 2015-07-31

Posts: 47

Re: Geometry Problem

Well, I guess they are the same, because it turned out to be right. Thanks again bob xD

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Geometry Problem

How about this?

In triangle ABD by cosine rule:

and in triangle ACD:

But those angles add up to 180 so

So adding

Bob

---

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