Finding area of triangle inside triangle

[mathisfun512]

Hello all,
I’m have a hard trying solving this problem. See attached problem and my work.
Maybe there’s an easier way but the only thing I could do is create a system of equations but it’s not helping.
Any help would be appreciated. Thank you!

 

[Dr.Peterson]

You're right, the system of equations is redundant, so you need additional information it doesn't capture.

One nice thing to do is to construct a line through N parallel to CM, and think about implications. I'm sure there are several other ways.

 

[mathisfun512]

Hi,
In doing so, I would create similar triangles. But using the similar triangles to set up proportions does not help me find the area in question. Maybe I’m missing something. Could you please point me in the right direction?
Thank you for your time.

 

[Dr.Peterson]

Suppose the new line meets AB at P. Consider triangles BCM and BNP, then triangles ANB and ANP, then triangles ANP and ATM.

 

[mathisfun512]

ANB and ANP are not similar triangles. You seem to be implying they are. Sorry, I am lost.

 

[Denis]

Well, there are no integer sided triangles with area 28;
closest are 4-13-15 and 6-8-10 with area 24.
So no "nice" solution!
Is the solution given?

 

[mathisfun512]

No solution given.

 

[Dr.Peterson]

The answer is reasonably "nice"; it's rational. Some of the other areas are integers. And early in my work I started seeing 7's, which told me the number 28 was chosen carefully.

 

[mathisfun512]

The system I created (see initial post):
x+y =21
y+w=14
x+z=14
z+w=7
Furthermore, based on the figure, we can tell that x>y>w>z
I started using trial and errors and came up with a solution of the system since the system can’t be solved using elimination or substitution.
The solution is:
x=13
y=9
w=5
z=2

So the area of triangle ATM is y =9.
Still looking for a better method.

 

[Dr.Peterson]

The system I created (see initial post):
x+y =21 [1]
y+w=14 [2]
x+z=14 [3]
z+w=7 [4]
Furthermore, based on the figure, we can tell that x>y>w>z
I started using trial and errors and came up with a solution of the system since the system can’t be solved using elimination or substitution.
The solution is:
x=13
y=9
w=5
z=2

So the area of triangle ATM is y =9.
Still looking for a better method.

How did you use trial and error? Why you do think you can be sure of your inequality by looking at the figure?

How do you know this is the answer to the problem, not just a solution to the system of equations? (I told you the system is redundant - by which I mean it has more than one solution, and isn't sufficient to determine the solution to the problem. It's in that sense that you "can't solve it".)

In fact, it is not a solution; your x+y = 22, not 21. Your area for ATM, however, happens to be correct. (I made a mistake earlier, doing the work in my head, and thought it was just a nice fraction.)

Here is the reality of the system of equations: We can eliminate w by subtracting [4] from [2], to get [5] y - z = 7. But adding this to [3] yields [1], so the system is dependent. (That is, you can drop an equation without losing any information.) We could take any value for w, and find that a solution will be z=7-w, y=14-w, and x=7+w. (Take w=5 and you get your answer, except that x=12.)

Suppose that w=4.5 instead. Then x=11.5, y=9.5, and z=2.5; that still fits your inequality, as well as all four equations. So you can't tell whether you are right.

Did you try following my suggestions? They will give you the better method you are looking for.

 

[mathisfun512]

How did you use trial and error? Why you do think you can be sure of your inequality by looking at the figure?

How do you know this is the answer to the problem, not just a solution to the system of equations? (I told you the system is redundant - by which I mean it has more than one solution, and isn't sufficient to determine the solution to the problem. It's in that sense that you "can't solve it".)

In fact, it is not a solution; your x+y = 22, not 21. Your area for ATM, however, happens to be correct.

Here is the reality of the system of equations: We can eliminate w by subtracting [4] from [2], to get [5] y - z = 7. But adding this to [3] yields [1], so the system is dependent. (That is, you can drop an equation without losing any information.) We could take any value for w, and find that a solution will be z=7-w, y=14-w, and x=7+w. (Take w=5 and you get your answer, except that x=12.)

Suppose that w=4.5 instead. Then x=11.5, y=9.5, and z=2.5; that still fits your inequality, as well as all four equations. So you can't tell whether you are right.

Did you try following my suggestions? They will give you the better method you are looking for.

I got my values carefully by studying the proportions. I am absolutely aware that the solution I got was not unique to the system. That’s why I said in my initial post that the system can’t be solved to find a unique solution. I was not absolutely randomly picking just any solution of the system. The proportions given by the problem make it possible to rank x, y, w, and z.

x+y = 21. I showed you how I got that. You’re saying it’s 22, can you show me how you got 22?

And no, I still can utilize the information you gave me because one of the pairs of triangles you implied were similar but they are not (the second pair you gave).

 

[Dr.Peterson]

I got my values carefully by studying the proportions. I am absolutely aware that the solution I got was not unique to the system. That’s why I said in my initial post that the system can’t be solved to find a unique solution. I was not absolutely randomly picking just any solution of the system. The proportions given by the problem make it possible to rank x, y, w, and z.

The problem can be solved. Your system of equations, as I said, doesn't contain all the information; the fact that it doesn't have a unique solution does not mean the problem doesn't have a unique solution. I also showed you that your solution is not unique even with the addition of the inequality.

Trial and error (systematic, not merely random) can be a valid solution method for some problems. It is not in this case, no matter how intelligently you did it, unless you used something you haven't mentioned to confirm that your solution is actually correct.

x+y = 21. I showed you how I got that. You’re saying it’s 22, can you show me how you got 22?

What I said was not that the equation was wrong, but that your claimed solution, with x=13 and y=9, did not satisfy that equation! You probably just mistyped and meant x=12. Right?

And no, I still can't utilize the information you gave me because one of the pairs of triangles you implied were similar but they are not (the second pair you gave).

I didn't say those particular triangles are similar. I said they can be used if you "consider" them in some unspecified way. (It looks like I missed your response #5 stating that you inferred that I meant ANB and ANP are similar, due to moderation delays; otherwise I might have said this sooner.)

Please TRY looking for ways to use my hints. The whole point of assigning this sort of problem is to teach you to think through non-routine problems, without having to be told every step to take.

But here's another hint: In addition to similar triangles, a common fact to use in this sort of area-ratio problem is that two triangles with the same base have areas proportional to their altitudes; another is that two triangles with the same altitude have areas proportional to their bases. Either of those can be used here. (There are probably lots of other ways to get through this problem.)

Finally, if you want efficient help, try showing us what you actually did, rather than just saying it didn't work. We can correct errors, or give more useful hints, if we know the direction you are going.

 

[mathisfun512]

The problem can be solved. Your system of equations, as I said, doesn't contain all the information; the fact that it doesn't have a unique solution does not mean the problem doesn't have a unique solution. I also showed you that your solution is not unique even with the addition of the inequality.

Trial and error (systematic, not merely random) can be a valid solution method for some problems. It is not in this case, no matter how intelligently you did it, unless you used something you haven't mentioned to confirm that your solution is actually correct.

What I said was not that the equation was wrong, but that your claimed solution, with x=13 and y=9, did not satisfy that equation! You probably just mistyped and meant x=12. Right?

I didn't say those particular triangles are similar. I said they can be used if you "consider" them in some unspecified way. (It looks like I missed your response #5 stating that you inferred that I meant ANB and ANP are similar, due to moderation delays; otherwise I might have said this sooner.)

Please TRY looking for ways to use my hints. The whole point of assigning this sort of problem is to teach you to think through non-routine problems, without having to be told every step to take.

But here's another hint: In addition to similar triangles, a common fact to use in this sort of area-ratio problem is that two triangles with the same base have areas proportional to their altitudes; another is that two triangles with the same altitude have areas proportional to their bases. Either of those can be used here. (There are probably lots of other ways to get through this problem.)

Finally, if you want efficient help, try showing us what you actually did, rather than just saying it didn't work. We can correct errors, or give more useful hints, if we know the direction you are going.

I never said the problem can’t be solved.
In fact, I solved it and your answer matches mine. I just wanted a better method as I stated in my previous post.
I obviously meant x =12. This was a typo.

You keep telling me that the system I setup doesn’t have a unique solution even after I’ve told you that I was fully aware that it doesn’t have a unique solution (I even mentioned in my very first post that my system can’t be solved uniquely). So I’m not sure why you keep telling me that the system doesn’t have a unique solution since I repeatedly stated so from the get go and you proceeded to come up with other solutions to show me it doesn’t have a unique solution even though I stated that there were other solutions but I came up with what I came up with using trials and careful errors.
Like I said twice already, I am fully aware that the system I setup doesn’t one only one solution, and also fully aware that the problem has a unique solution which is what my answer tried to show (the solution I posted).

I did show all my work, a full page of work that is. I’m not one to just say “how do you solve this?”.

 

[Dr.Peterson]

Let's not argue over who said what when.

The goal is for you to find the better solution, for which I gave hints. You haven't shown me any of the work USING THAT METHOD that I asked you to show. Please do. Let's move forward.

 

[mathisfun512]

Let's not argue over who said what when.

The goal is for you to find the better solution, for which I gave hints. You haven't shown me any of the work USING THAT METHOD that I asked you to show. Please do. Let's move forward.

I told you that I don’t see it. There is NO work to be shown when someone says they don’t see it.
You had a condescending tone which put me on the defensive. I no longer want to pursue this discussion.
Thanks.

 

[Dr.Peterson]

I told you that I don’t see it. There is NO work to be shown when someone says they don’t see it.
You had a condescending tone which put me on the defensive. I no longer want to pursue this discussion.
Thanks.

If you should choose to try, just show me whatever work you did see to do.

You said that one pair of triangles were not similar, so you must have seen some things that you could do. And I gave you an additional idea for the pair you didn't see.

But, yes, you are allowed to end the discussion.

 

[Dr.Peterson]

But, of course, it's a very nice problem, and others may want to see how to solve it; so let me offer the further thoughts I intended to give eventually.

Here is a picture:



I said this:

Suppose the new line meets AB at P. Consider triangles BCM and BNP, then triangles ANB and ANP, then triangles ANP and ATM.

First, triangles BCM and BNP are similar. Knowing that BN : BC = 2:1, what is BP:BM? Then, what is BP : BA? (Alternatively, you could use this pair to find area(BNP).)

Next, ANB and ANP have the same altitude and different bases. What is the ratio of their areas?

Finally, ANP and ATM are similar. Knowing MA : PA, what is the ratio of their areas?