B More similar triangle problems

[DaveC426913]

TL;DR Summary

Is this solution based on a valid postulate?

For the sake of argument, lets assume this is, in fact, a right triangle with a rectangle inscribed in it.

Here is what looks like a very elegant solution:

Triangles A and B are similar
a/4=2/b
ab=8

But what is happening in that second line? Is there a property of similar triangles I'm forgetting?

 

[fresh_42]

What do you mean? Similarity causes the first line, and the second is just a multiplication by $4\cdot b.$ Are you asking why they are similar?

 

[Mark44]

But what is happening in that second line? Is there a property of similar triangles I'm forgetting?

Similar triangles have corresponding sides that are in the same proportion. Therefore, a : 4 :: 2 : b, or equivalently, $\frac a 4 = \frac 2 b$.

I wouldn't call ab = 8 a solution, though, as you haven't solved for either a or b. You can solve for a and b by computing the areas of the two internal triangles and the rectangle, and equating the sum of those areas with the area of the large triangle.

 

[TensorCalculus]

$$\frac a 4 = \frac 2 b$$
$$a = 4(\frac 2 b)$$
$$ab = 4(2) = 8$$

How $ab=8$ is a full solution I'm not sure, but hopefully laying everything out there makes everything clear.

 

[DaveC426913]

Sorry. I left it off. The question is what is the area of the shaded region?

Which is why the problem is solvable.
There is not enough info to know the length of sides of the shaded area, but its area is constant.

 

[DaveC426913]

It was simply the arrival at this line:

$$\frac a 4 = \frac 2 b$$

I can see it now, but I never would have thought of it in my own.

 

[fresh_42]

Here is the long version:

The area of the rectangle is $(2+a)\cdot(4+b) =8+4a+2b+ab,$ so half of it is the area of the triangle. The triangle also consists of the colored areas. The yellow area is $b,$ the blue one is $2a,$ and the pink one is $ab$. So
$$
\dfrac{1}{2}\left(8+4a+2b+ab\right)=4+2a+b+\dfrac{1}{2}ab=b+2a+ab\Longleftrightarrow 4=\dfrac{1}{2}ab\Longleftrightarrow ab=8
$$
This shows that the similarity argument is indeed the nicer one.

 

[Mark44]

There is not enough info to know the length of sides of the shaded area

Sure there is. I laid out how to find the dimensions of the rectangle in post #3. Equating the areas as I described gives you a second equation in a and b, allowing one to solve for both variables.

 

[PAllen]

Sure there is. I laid out how to find the dimensions of the rectangle in post #3. Equating the areas as I described gives you a second equation in a and b, allowing one to solve for both variables.

Write out your method and you will see you get an identity. Same for computing each triangle hypotenuse and relating them. The issue is that the problem is fundamentally undetermined as to a and b, as long as their product is 8. You can see this directly by graphically showing that e.g. 1 and 8, 2 and 4, both work.

That’s why I thought this was a cute problem. Similar to the case where you drill a hole through a sphere. No matter the sphere size, if you make the hole wide enough that the height of the resulting figure is e.g. 8, the volume of what’s left is independent of sphere size.

 

[DaveC426913]

Sure there is. I laid out how to find the dimensions of the rectangle in post #3. Equating the areas as I described gives you a second equation in a and b, allowing one to solve for both variables.

Right. I must have misintuited the problem. At first, I was thinking the ratio of sides of the rectangle were undefined - that the rectangle could change shape but not area.

 

[PAllen]

Right. I must have misintuited the problem. At first, I was thinking the ratio of sides of the rectangle were undefined - that the rectangle could change shape but not area.

Your first instinct is correct.

 

[PAllen]

Exercise for the reader: calling the 4 and 2 segments c and d, show that for a given rectangle area A, any choice of a and b, with ab=A, and independently, any choice of c and d with cd=A, leads to a valid instance of rectangle inscribed in a right triangle.

 

[Mark44]

At first, I was thinking the ratio of sides of the rectangle were undefined - that the rectangle could change shape but not area.

If all you have is the relationship ab = 8 for the rectangle, then you can't get a unique solution for the sides a and b. But as I laid out, developing another equation nails down the dimensions of the rectangle.

 

[Mark44]

At first, I was thinking the ratio of sides of the rectangle were undefined - that the rectangle could change shape but not area.

If all you have is the relationship ab = 8 for the rectangle, then you can't get a unique solution for the sides a and b. But as I laid out, developing another equation nails down the dimensions of the rectangle.

 

[Mark44]

Write out your method and you will see you get an identity.

I wrote out my two equations but did not get an identity. I ended up with solutions of a = 4 and b = 2. I checked my work and found that these numbers satisfied both equations; i.e., that ab = 8 and that the sum of the three areas was 18.

If I made a mistake, I'm away from the house until tomorrow and am without paper and pen, so unable to recreate my equations.

 

[PAllen]

I wrote out my two equations but did not get an identity. I ended up with solutions of a = 4 and b = 2. I checked my work and found that these numbers satisfied both equations; i.e., that ab = 8 and that the sum of the three areas was 18.

If I made a mistake, I'm away from the house until tomorrow and am without paper and pen, so unable to recreate my equations.

Here it is worked out:

Sum of small triangles plus rectangle = large triangle

$$\frac {1}{2}4a + \frac{1}{2}2b + ab = \frac{1}{2}(4+b)(2+a)$$

simplify a little:

$$2a+b+ab=4+b+2a+ab/2$$

Substitute $b=8/a$ and simplify:

$$2a+8/a+8=8+8/a+2a$$

an identity, leading to $0=0$.

Using big hypotenuse = sum of smaller hypotenuses also leads to an identity. There is no more information, because the exercise I posted in #12 is straightforward to prove.

 

[Mark44]

I don't have my work in front of me, but it appears I've made an error.