Triangle formed by a plane intersecting the coordinate axis.

Question

T is the region of the plane $x+y+z=1,x,y,z>0$. S is the set points $(a,b,c)$ in T such that just two of the following three inequalities hold: $a\le\frac12,b\le\frac13,c\le\frac16$. Find area of the region S.

T is an equilateral triangle with vertices $(0,0,1),(0,1,0),(1,0,0)$. Its side is $\sqrt{2}$

So, area of T=$\frac{√3}4(√2)^2=\frac{\sqrt3}2$

I am not able to visualise S.

On Byju's website, they have subtracted areas of three small equilateral triangles. I am not able to visualise that.

On AoPS website, they have drawn a colorful diagram but alas, I am still not able to visualise which three triangles are being talked about.

Is there any website or tool which could help see such 3D structures?

Or, if you could help me visualise it with better labelling or more expository language, I'll be grateful.

(On Byjus and AoPS, inequalities signs are opposite. I am not sure if both are same questions or different.)

Also, how have they taken the side lengths for small triangles?

EDIT:

Rephrasing my question:

1. Are the questions on Byjus and AoPS websites the same? If no, then how to solve the question that I have posted? If yes, then which of them has a typo?

2. How to find the sides of the small equilateral triangles?

3. How to visualise two inequalities at the same time?

Edit$2$:

On AoPS website, they have inequalities $x\ge a, y\ge b, z\ge c$.

S represents area where exactly two inequalities hold. e.g. the case $x\ge a, y\ge b, z\le c$.

If we subtract this area from total area, we would get cases like $x\ge a, y\le b, z\le c$. And $x\le a, y\le b, z\le c$. That means the subtraction doesn't really represent the question I have posted. Or does it?

Answer 1

If you find it difficult to consider the inequalities on the plane $x+y+z=1$, then I think fixing one of the variables and considering the orthogonal projection should help.

Let us fix $z=k$ where $k$ is a real number such that $0<k<1$. Then, we see that $\{x,y\mid x+y+k=1,x>0,y>0,z=k\}$, i.e. $\{x,y\mid y=-x+1-k,0<x<1-k,z=k\}$ represents a line segment (except the endpoints) on the plane $z=k$.

Examples :

• $\{x,y\mid y=-x+1,0<x<1,z=0\}$ represents a line segment on the plane $z=0$ (though we have to have $z\gt 0$).

• $\{x,y\mid y=-x+\frac 56,0<x<\frac 56,z=\frac 16\}$ represents a line segment on the plane $z=\frac 16$.

• $\{x,y\mid y=-x+\frac 12,0<x<\frac 12,z=\frac 12\}$ represents a line segment on the plane $z=\frac 12$.

• $\{x,y\mid y=-x+\frac 13,0<x<\frac 13,z=\frac 23\}$ represents a line segment on the plane $z=\frac 23$.

• $\{x,y\mid y=-x,x=0,z=1\}$ represents a point on the plane $z=1$ (though we have to have $x\gt 0$).

Now, let us consider the orthogonal projection of these line segments onto the $xy$ plane.

Figure 1 :

• $DH$ is the line segment $\{x,y\mid y=-x+1,z=0\}$ on the $xy$ plane where $O$ is the origin, and $D$ is on the $x$-axis, and $H$ is on the $y$-axis.

• $CG$ is the orthogonal projection of the line segment $\{x,y\mid y=-x+\frac 56,0<x<\frac 56,z=\frac 16\}$ onto the $xy$ plane.

• $BF$ is the orthogonal projection of the line segment $\{x,y\mid y=-x+\frac 12,0<x<\frac 12,z=\frac 12\}$ onto the $xy$ plane.

• $AE$ is the orthogonal projection of the line segment $\{x,y\mid y=-x+\frac 13,0<x<\frac 13,z=\frac 23\}$ onto the $xy$ plane.

• $O$ is the orthogonal projection of the point $\{x,y\mid y=-x,x=0,z=1\}$ onto the $xy$ plane.

Now, let us consider the following three cases :

case 1 : $a+b+c=1,a\le\frac 12,b\le\frac 13,c\gt \frac 16$

case 2 : $a+b+c=1,a\le\frac 12,b\gt\frac 13,c\le \frac 16$

case 3 : $a+b+c=1,a\gt\frac 12,b\le\frac 13,c\le\frac 16$

Considering the orthogonal projection of such $(a,b,c)$ onto the $xy$ plane, we can "visualize" such $(a,b,c)$ as follows :

Figure 2 :

Explanations :

• $a\le\frac 12$ represents the left region of the line $BJ$.

• $a\gt\frac 12$ represents the right region of the line $BJ$.

• $b\le\frac 13$ represents the region below the line $EK$.

• $b\gt \frac 13$ represents the region above the line $EK$.

• $c\le\frac 16$ represents the region above the line $CG$.

• $c\gt \frac 16$ represents the region below the line $CG$.

• $a\le\frac 12,b\le\frac 13,c\gt\frac 16$ represents the quadrilateral $OBIE$.

• $a\le\frac 12,b\gt\frac 13,c\le\frac 16$ represents the quadrilateral $IJHG$.

• $a\gt\frac 12,b\le\frac 13,c\le\frac 16$ represents the quadrilateral $CIKD$.

Since Figure 2 represents the orthogonal projection of $x+y+z=1$ with some inequalities onto the $xy$ plane, from here, let us finally consider the plane $x+y+z=1$ itself.

Figure 3 :

Here, $\triangle{IBC},\triangle{IEG},\triangle{IKJ}$ are equilateral triangles with side lengths $\frac{\sqrt 2}{3},\frac{\sqrt 2}{2},\frac{\sqrt 2}{6}$ respectively. (This is because $\frac{BC}{OD}=\frac{2}{6}=\frac 13,\frac{EG}{OH}=\frac{3}{6}=\frac 12$ and $\frac{KJ}{DH}=\frac 16$.)

Therefore, we see that the area of $S$ is given by $$S=\frac{\sqrt 3}{2}-\frac{\sqrt 3}{4}\bigg(\bigg(\frac{\sqrt 2}{3}\bigg)^2+\bigg(\frac{\sqrt 2}{2}\bigg)^2+\bigg(\frac{\sqrt 2}{6}\bigg)^2\bigg)=\color{red}{\frac{11}{36}\sqrt 3}$$

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Are the questions on Byjus and AoPS websites the same? If no, then how to solve the question that I have posted? If yes, then which of them has a typo?

The question on Byju's website is the same as yours (though it says $a\color{red}{\lt}\frac 12$, maybe typo?), but I think the answer is wrong. The answer says that the side lengths of small equilateral triangles are $\frac{1}{2},\frac{1}{3},\frac{1}{6}$. I think this is wrong.

The question on AoPS website is not the same as yours. The question on AoPS is about "exactly two of $a\color{red}{\ge}\frac12,b\color{red}{\ge}\frac13,c\color{red}{\ge}\frac16$ hold".

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Added :

I understand $CG$ is the line segment on the plane $z=\frac 16$. But how does $c\le\frac 16$ represent the region above $CG$? Shouldn't it be below $CG$?

If you understand $CG$ is the orthogonal projection of the line segment on the plane $z=\frac 16$, then you should be able to understand $BF$ is the orthogonal projection of the line segment on the plane $z=\frac 12$ which is bigger than $\frac 16$.

Since we are considering the line segment of the form $y=-x+1-k,0<x<1-k$ on $z=k$, let us consider the $y$-intercept $1-k$.

You'll see that the $y$-intercept $1-k$ is decreasing from $1$ (when $k=0$) to $0$ (when $k=1$). So, the bigger $k$ is, the smaller the $y$-intercept is. The smaller $k$ is, the bigger the $y$-intercept is.

This implies that $c\le \frac 16$ represents the region above $CG$.

Answer 2

I have marked up the portion in the first image. I hope now you can understand which three triangles they were referring.

Edit:

1)I am not sure about Byju's solution but I could understand AoPs solution so I refer to it. The area required in the AoPs problem was for the opposite of your inequalities (i.e. they were for $\ge$ but you need $\le$). To find the area you need, you can subtract area of three highlighted triangles from area of $S$ to get you required area.

2)To get the side lengths of the three equilateral triangles you need to solve that equations of the planes $x=\frac12,y=\frac13,z=\frac16$ along with $x+y+z=1$ to get point of intersection from where you can find length of the sides from the vertices of the bigger triangle using distance formula.

3)Try to use a 3D graph plotter, for better visualization (Here's one:https://www.math3d.org/) and try to play around.

Answer to Edit 2:

Can you point out in the image where we will get the area $x \le \dfrac 12,y \le \dfrac 13,z \le \dfrac 16$ on the plane $x+y+z=1$.

Please try to understand that the area we are concerned with needs to lie on the plane $x+y+z=1$. Thus we would never get a area on the plane where $x \le \dfrac 12,y \le \dfrac 13,z \le \dfrac 16$ simultaneously since $x+y+z \lt 1$ and thus the point would lie outside the plane. Thus subtracting the area of $S$ from total area would represent the area of your question.