A triangle is divided by two cevians into four regions of integer areas $A,B,C,D$.

What are the possible values of the area of the triangle ($A+B+C+D$)?

Expressing the question in terms of an equation

To find the relationship among $A,B,C,D$, split the region with area $C$ into regions of areas $x$ and $y$, as shown.

Considering the triangles with base along the red cevian gives $\frac{A}{D}=\frac{B+y}{x}$.

Considering the triangles with base along the green cevian gives $\frac{A}{B}=\frac{D+x}{y}$.

Solving for $x$ and $y$ gives $x=\frac{ABD+BD^2}{A^2-BD}$ and $y=\frac{ABD+B^2D}{A^2-BD}$, so $C=x+y=\frac{(2A+B+D)BD}{A^2-BD}$.

So my question is:

If $C=\frac{(2A+B+D)BD}{A^2-BD}$ where $A,B,C,D\in\mathbb{Z}^+$, what are the possible values of $A+B+C+D$?

Context

Every New Year, I give my students a geometry puzzle like the following.

"The numbers in the (not-to-scale) diagram below are areas. What is the total area of the triangle?"

The area of the unmarked region must be $1848$, so the area of the triangle is $54+96+1848+27=2025$.

I began doing this in $2020$, and every year since then I have been able to create such a puzzle. That is, I could always find (using Excel) four integer areas such that their sum equals the number of the new year:

• $180+729+1089+22=2020$
• $94+94+1755+78=2021$
• $674+337+674+337=2022$
• $289+578+1088+68=2023$
• $46+46+1890+42=2024$
• $54+96+1848+27=2025$

But I have found that it is impossible to create such a puzzle for $2026$, after an exhaustive search.

Now I wonder, what are the possible areas of the triangle? Do they follow a simple pattern?

Edit

I found that OEIS sequence A208770 lists the smallest $60$ possible areas of the triangle, but gives no insight about the pattern.

This sequence is very similar to the first $60$ish terms of two other OEIS sequences:

• A278638, which lists numbers $n$ such that $1/n$ is a difference of Egyptian fractions with all denominators less than $n$. A278638 contains three terms ($33,68,76$) not found in A208770.
• A219095 which lists numbers $k$ such that if $b/c>1$ is the least reduced fraction using divisors $b$ and $c$ of $k$, then $c>1$). A219095 is missing three terms ($44,52,136$) found in A208770.

values of $A+B+C+D$ if $C=\frac{(2A+B+D)BD}{A^2-BD}$

A triangle is divided by two cevians into four regions of integer areas $A,B,C,D$.

What are the possible values of the area of the triangle ($A+B+C+D$)?

Expressing the question in terms of an equation

To find the relationship among $A,B,C,D$, split the region with area $C$ into regions of areas $x$ and $y$, as shown.

Considering the triangles with base along the red cevian gives $\frac{A}{D}=\frac{B+y}{x}$.

Considering the triangles with base along the green cevian gives $\frac{A}{B}=\frac{D+x}{y}$.

Solving for $x$ and $y$ gives $x=\frac{ABD+BD^2}{A^2-BD}$ and $y=\frac{ABD+B^2D}{A^2-BD}$, so $C=x+y=\frac{(2A+B+D)BD}{A^2-BD}$.

So my question is:

If $C=\frac{(2A+B+D)BD}{A^2-BD}$ where $A,B,C,D\in\mathbb{Z}^+$, what are the possible values of $A+B+C+D$?

Context

Every New Year, I give my students a geometry puzzle like the following.

"The numbers in the (not-to-scale) diagram below are areas. What is the total area of the triangle?"

The area of the unmarked region must be $1848$, so the area of the triangle is $54+96+1848+27=2025$.

I began doing this in $2020$, and every year since then I have been able to create such a puzzle. That is, I could always find (using Excel) four integer areas such that their sum equals the number of the new year:

• $180+729+1089+22=2020$
• $94+94+1755+78=2021$
• $674+337+674+337=2022$
• $289+578+1088+68=2023$
• $46+46+1890+42=2024$
• $54+96+1848+27=2025$

But I have found that it is impossible to create such a puzzle for $2026$, after an exhaustive search.

Now I wonder, what are the possible areas of the triangle? Do they follow a simple pattern?

Edit

I found that OEIS sequence A208770 lists the smallest $60$ possible areas of the triangle, but gives no insight about the pattern.

This sequence is very similar to the first $60$ish terms of two other OEIS sequences:

• A278638, which lists numbers $n$ such that $1/n$ is a difference of Egyptian fractions with all denominators less than $n$. A278638 contains three terms ($33,68,76$) not found in A208770.
• A219095 which lists numbers $k$ such that if $b/c>1$ is the least reduced fraction using divisors $b$ and $c$ of $k$, then $c>1$). A219095 is missing three terms ($44,52,136$) found in A208770.

A triangle is divided by two cevians into four regions of integer areas $A,B,C,D$.

What are the possible values of the area of the triangle ($A+B+C+D$)?

Expressing the question in terms of an equation

To find the relationship among $A,B,C,D$, split the region with area $C$ into regions of areas $x$ and $y$, as shown.

Considering the triangles with base along the red cevian gives $\frac{A}{D}=\frac{B+y}{x}$.

Considering the triangles with base along the green cevian gives $\frac{A}{B}=\frac{D+x}{y}$.

Solving for $x$ and $y$ gives $x=\frac{ABD+BD^2}{A^2-BD}$ and $y=\frac{ABD+B^2D}{A^2-BD}$, so $C=x+y=\frac{(2A+B+D)BD}{A^2-BD}$.

So my question is:

If $C=\frac{(2A+B+D)BD}{A^2-BD}$ where $A,B,C,D\in\mathbb{Z}^+$, what are the possible values of $A+B+C+D$?

Context

Every New Year, I give my students a geometry puzzle like the following.

"The numbers in the (not-to-scale) diagram below are areas. What is the total area of the triangle?"

The area of the unmarked region must be $1848$, so the area of the triangle is $54+96+1848+27=2025$.

I began doing this in $2020$, and every year since then I have been able to create such a puzzle. That is, I could always find (using Excel) four integer areas such that their sum equals the number of the new year:

• $180+729+1089+22=2020$
• $94+94+1755+78=2021$
• $674+337+674+337=2022$
• $289+578+1088+68=2023$
• $46+46+1890+42=2024$
• $54+96+1848+27=2025$

But I have found that it is impossible to create such a puzzle for $2026$, after an exhaustive search.

Now I wonder, what are the possible areas of the triangle? Do they follow a simple pattern?

A triangle is divided by two cevians into four regions of integer areas $A,B,C,D$.

What are the possible values of the area of the triangle ($A+B+C+D$)?

Expressing the question in terms of an equation

To find the relationship among $A,B,C,D$, split the region with area $C$ into regions of areas $x$ and $y$, as shown.

Considering the triangles with base along the red cevian gives $\frac{A}{D}=\frac{B+y}{x}$.

Considering the triangles with base along the green cevian gives $\frac{A}{B}=\frac{D+x}{y}$.

Solving for $x$ and $y$ gives $x=\frac{ABD+BD^2}{A^2-BD}$ and $y=\frac{ABD+B^2D}{A^2-BD}$, so $C=x+y=\frac{(2A+B+D)BD}{A^2-BD}$.

So my question is:

If $C=\frac{(2A+B+D)BD}{A^2-BD}$ where $A,B,C,D\in\mathbb{Z}^+$, what are the possible values of $A+B+C+D$?

Context

Every New Year, I give my students a geometry puzzle like the following.

"The numbers in the (not-to-scale) diagram below are areas. What is the total area of the triangle?"

The area of the unmarked region must be $1848$, so the area of the triangle is $54+96+1848+27=2025$.

I began doing this in $2020$, and every year since then I have been able to create such a puzzle. That is, I could always find (using Excel) four integer areas such that their sum equals the number of the new year:

• $180+729+1089+22=2020$
• $94+94+1755+78=2021$
• $674+337+674+337=2022$
• $289+578+1088+68=2023$
• $46+46+1890+42=2024$
• $54+96+1848+27=2025$

But I have found that it is impossible to create such a puzzle for $2026$, after an exhaustive search.

Now I wonder, what are the possible areas of the triangle? Do they follow a simple pattern?

Edit

I found that OEIS sequence A208770 lists the smallest $60$ possible areas of the triangle, but gives no insight about the pattern.

This sequence is very similar to the first $60$ish terms of two other OEIS sequences:

• A278638, which lists numbers $n$ such that $1/n$ is a difference of Egyptian fractions with all denominators less than $n$. A278638 contains three terms ($33,68,76$) not found in A208770.
• A219095 which lists numbers $k$ such that if $b/c>1$ is the least reduced fraction using divisors $b$ and $c$ of $k$, then $c>1$). A219095 is missing three terms ($44,52,136$) found in A208770.

A triangle is divided by two cevians into four regions of integer areas $A,B,C,D$.

What are the possible values of the area of the triangle ($A+B+C+D$)?

Expressing the question in terms of an equation

Two find the relationship among $A,B,C,D$, split the region with area $C$ into regions of areas $x$ and $y$, as shown.

Considering the triangles with base along the red cevian gives $\frac{A}{D}=\frac{B+y}{x}$.

Considering the triangles with base along the green cevian gives $\frac{A}{B}=\frac{D+x}{y}$.

Solving for $x$ and $y$ gives $x=\frac{ABD+BD^2}{A^2-BD}$ and $y=\frac{ABD+B^2D}{A^2-BD}$, so $C=x+y=\frac{(2A+B+D)BD}{A^2-BD}$.

So my question is:

Given $C=\frac{(2A+B+D)BD}{A^2-BD}$ where $A,B,C,D\in\mathbb{Z}^+$, what are the possible values of $A+B+C+D$?

Context

Every New Year, I give my students a geometry puzzle like the following.

"The numbers in the diagram below are areas. What is the total area of the triangle?"

The area of the unmarked region must be $1848$, so the area of the triangle is $54+96+1848+27=2025$.

I began doing this in $2020$, and every year since then I have been able to create such a puzzle. That is, I could always find (using Excel) four integer areas such that their sum equals the number of the new year:

• $180+729+1089+22=2020$
• $94+94+1755+78=2021$
• $674+337+674+337=2022$
• $289+578+1088+68=2023$
• $46+46+1890+42=2024$
• $54+96+1848+27=2025$

But I have found that it is impossible to create such a puzzle for $2026$, after an exhaustive search. So my holiday tradition cannot continue.

Now I wonder, what are the possible areas of the triangle? Do they follow a simple pattern?

A triangle is divided by two cevians into four regions of integer areas $A,B,C,D$.

What are the possible values of the area of the triangle ($A+B+C+D$)?

Expressing the question in terms of an equation

To find the relationship among $A,B,C,D$, split the region with area $C$ into regions of areas $x$ and $y$, as shown.

Considering the triangles with base along the red cevian gives $\frac{A}{D}=\frac{B+y}{x}$.

Considering the triangles with base along the green cevian gives $\frac{A}{B}=\frac{D+x}{y}$.

Solving for $x$ and $y$ gives $x=\frac{ABD+BD^2}{A^2-BD}$ and $y=\frac{ABD+B^2D}{A^2-BD}$, so $C=x+y=\frac{(2A+B+D)BD}{A^2-BD}$.

So my question is:

If $C=\frac{(2A+B+D)BD}{A^2-BD}$ where $A,B,C,D\in\mathbb{Z}^+$, what are the possible values of $A+B+C+D$?

Context

Every New Year, I give my students a geometry puzzle like the following.

"The numbers in the (not-to-scale) diagram below are areas. What is the total area of the triangle?"

The area of the unmarked region must be $1848$, so the area of the triangle is $54+96+1848+27=2025$.

I began doing this in $2020$, and every year since then I have been able to create such a puzzle. That is, I could always find (using Excel) four integer areas such that their sum equals the number of the new year:

• $180+729+1089+22=2020$
• $94+94+1755+78=2021$
• $674+337+674+337=2022$
• $289+578+1088+68=2023$
• $46+46+1890+42=2024$
• $54+96+1848+27=2025$

But I have found that it is impossible to create such a puzzle for $2026$, after an exhaustive search.

Now I wonder, what are the possible areas of the triangle? Do they follow a simple pattern?