3 Rectangles + 1 Triangle = ?

Let the side lengths of $\triangle ABC$ be $a$ , $b$ , and $c$ . Then, we have: $a(b + c) + b(a + c) = c(a + b)^2,$ from our conditions in the problem. We expand each side and simplify as follows: $2ab + ac + bc = (a^2)c + (b^2)c + 2abc$ $2ab(c - 1) + ac(a - 1) + bc(b - 1) = 0$ Since $a \ne 0, b \ne 0$ , and $c \ne 0$ , we have that $a, b, c \ge 1$ . So: $2ab(c - 1) \ge 0, bc(b - 1) \ge 0, ac(c - 1) \ge 0$ We want the equality case; i.e. $a = b = c = 1.$ Thus, there is only this one possibility, and equilateral triangle with area $\frac {\sqrt{3} \cdot 1^2}{4}.$ The answer is $\boxed {1}$ .

Let $a = BC, b = AC, c = AB$ . Then the $[ABHJ]=[BCDE]+[ACFG]$ becomes $c(b+a)^2=a(b+c)+b(a+c)$ , which simplifies to $bc(b-1)+ac(a-1)+2ab(c-1)=0$ Now using the fact that $a\geq1, b\geq1, c\geq1, a,b,c \in \mathbb{Z}$ Each part of the left hand side is non-negative. So it can only sum to zero if each of them is 0. Thus, the sides are $a = b = c = 1$ .

Let $BC=a, AC=b, AB=c$ where $a,b,c \in \mathbb{N}$ $\;\;\;$

then $CD= b+c, CF= c+a, BH= (a+b)^2$ $\;\;\;$

so, $[ABHJ] = c(a+b)^2$ , $[BCDE] = a(b+c)$ , $[ACFG] =b(a+c)$

then we have

$c(a+b)^2 = a(b+c) +b(a+c)$

$a^2c + 2abc + b^2c = ab+ac +ab+bc$

$ac(a-1) + 2ab(c-1) + bc(b-1) = 0$

we can conclude that there is only one solution $(a,b,c) = (1,1,1)$

because of $a,b,c \geq 1$ so $ac(a-1),2ab(c-1),bc(b-1) \geq 0$ and the equality holds only for $(a,b,c) = (1,1,1)$

the different value of $[ABC]$ only $1$

Let $BC=a,AC=b,AB=c$

then $CD=b+c,CF=c+a,BH=(a+b)^2$

so, $[ABHJ]=c(a+b)^2, [BCDE]=a(b+c) , [ACFG]=b(a+c)$

then we have

$c(a+b)2=a(b+c)+b(a+c)$

$ac(a-1)+2ab(c-1)+bc(b-1)=0$

so that there is only one solution (a,b,c)=(1,1,1)

because of a,b,c≥1 so ac(a−1),2ab(c−1),bc(b−1)≥0 and the equality holds only for (a,b,c)=(1,1,1)

the different value of [ABC] only 1

Let the side opposite to $\angle A, \angle B$ and $\angle C$ be $a, b$ and $c$ respectively.

Given:

$CD=AC+AB=b+c$

$CF=AB+BC=c+a$

$BH=(AC+BC)^{2}=(a+b)^{2}$

$[ABHJ]=[BCDE]+[ACFG] \Rightarrow (a+b)^{2} \cdot c=a(b+c) +b(c+a)$

Expanding out, we have

$a^{2}c + 2abc + b^{2}c = ac+ 2ab + bc$

Rearranging and grouping like terms we have

$ac(a-1) +bc(b-1) + 2ab(c-1) = 0$

Also $a>0, b>0, c>0$

Since the triangle has integer side lengths the only solution possible is $a=1, b=1, c=1$

But since $a, b, c$ are sides of the triangle they can not be $0$

So the only possible solution is $a=b=c=1$

So only 1 triangle is possible.

Since the area of a rectangle is base times height, we know that $[ABHJ] = [BCDE] + [ACFG]$ becomes $BC(AC+AB) + AC(AB+BC) = AB(AC+BC)^2$ . Replacing $A = BC, B = AC, C = AB$ , we get $A(B+C) + B(C+A) = C(A+B)^2 = CA^2 + 2CAB + CB^2\ \Rightarrow$ $A(C+B-CA-CB) + B(A+C-CA-CB) = 0$
Since we know that A and B are positive, $C+B-CA-CB = A+C-CA-CB = 0 \Rightarrow A = B$ Plugging in we get $C + A = 2CA \Rightarrow (2C-1)(A-1/2) = 1/2$ $A = 1$ (for C to be a positive integer), $B = 1$ , and $C = 1$ . This is the only solution, so there is only one possible value of $[ABC]$ .

$[ABHJ] = [BCDE] + [ACFG]$ using formula area of rectangle i.e. length times width, we have; $AB(AC+BC)^2 = BC(AB+AC) + AC(AB+BC)$ $\Leftrightarrow AB(AC^2 + BC^2 + 2AC \cdot BC )$ $= BC(AB+AC) + AC(AB+BC)$

$\Leftrightarrow AB \cdot AC^2 + AB \cdot BC^2 + 2AB \cdot AC \cdot BC$ $= BC \cdot AB+BC \cdot AC) + AC \cdot AB + AC \cdot BC$

$\Leftrightarrow AB \cdot AC^2 - AB \cdot AC + AB \cdot BC^2 - AB \cdot BC$ $+ 2AB \cdot AC \cdot BC - 2BC \cdot AC =0$

$\Leftrightarrow AB \cdot AC(AC - 1) + AB \cdot BC( BC - 1) + 2BC \cdot AC(AB - 1) =0 ...(@)$

Since the Triangle ABC has integer side then $AB, BC, AC \geq 1$ (length of side ABC cann't be 0). Hence

$AB \cdot AC(AC - 1) \geq 0$

$AB \cdot BC( BC - 1) \geq 0$

$2BC \cdot AC(AB - 1) \geq 0$

@ will be satisfy if this three inequlity hold them equality and the equality hold when $AB = BC = AC = 1$

the triangle that satisfy the condition is only the equilateral triangle with 1 unit side

so, the only value for $[ABC]$ is $\sqrt{3}/4$ or only one value for $[ABC]$ that satisfy the condition.

a(b + c) + b(a + c) = c(a + b)^2 2ab + ac + bc = (a^2)c + (b^2)c + 2abc 2ab(c - 1) + ac(a - 1) + bc(b - 1) = 0

We know that none of a, b, c can be 0 => a, b, c ≥ 1 => 2ab(c - 1) ≥ 0, bc(b - 1) ≥ 0 and ac(c - 1) ≥ 0

Equality holds only if a = b = c = 1

So, ABC is an equilateral triangle with sides 1

=> [ABC] = (√3)/4

Area of ABHJ = area of BCDE+area of ACFG

AB . BH=BC . CD + AC . CF

Since BH=(AC+BC)^2 , CD=AC+AB and CF=AB+BC we can say

AB . (AC+BC)^2 =BC . (AC+AB) + AC . (AB+BC)

AB. (AC^2+BC^2+2AC.BC)=BC.AC+BC.AB+AC.AB+AC.BC

for easy AB=x, BC=y, AC=z

xz^2+xy^2+2xyz-yz-xy-xz-yz=0

(xz^2-xz)+(xy^2-xy)+2 (xyz-yz)=0

xz (z-1) +xy (y-1)+2yz (x-1)=0

Triangle has positive integer side So each term of lhs would be zero We get x=y=z=1

Triangle are root(3)/4 Only 1 answer possible

Let the side opposite to vertex A be a , vertex B be b and that to vertex C be c. Now label all the sides of the triangle along with the sides if the rectangles as stated in the question . Now, we shall equate the areas as directed above. We get, $c(a+b)^2=a(c+b) + (a+c)b$ . Divide the whole equation by c , $(a+b)^2=a+(2ab)/c+b$ . Expanding the square , by the identity $(a+b))^2=a^2+b^2+2ab, a^2+b^2+2ab=a+(2ab)/c+b.$ Now, we take the like terms together to the L.H.S. $a(a-1)+b(b-1)+2ab(1-1/c)=0$ Now , as we know that a,b and c are positive as they are the sides of a triangle, we can see that each individual term should be equal to zero in order the final result to be zero. Taking each term 0 we get a=1, b=1 and c=1. Therefore , we can get only 1 such pair.

Let $AB=x, AC=y, BC=z$ . Then, since the problem tells us that $[ABHJ]+[BCDE]=[ACFG]$ , we have $z(x+y)+y(x+z)=x(y+z)^2$ . Expanding and rearranging gives us $xy^2+2yz+z^2-xz-2yz-xy=0$ . Rearranging this equation then gives us $x(y^2+z^2-y-z)+2(xyz-yz)=0$ . Since $x,y,z$ are sides of a triangle, clearly both of the parts being added are equal to 0. We can quickly see that the only way that this can happen is if $x,y,z$ are 0 or 1, but since they are positive integers we have $x=y=z=1$ . Thus, there is $\boxed{1}$ total possibility for the area of triangle ABC.

$With\ usual\ notation,\ \ [BJHA]=BCDE]+[ACFG]\\ \implies c\{a+b\}^2=a\{b+c\}+b\{c+a\}=2ab+c\{a+b\}.\\ So\ c\{a^2+b^2\}+2abc=2ab+c\{a+b\} .\\ \implies \ c\{ a^2+b^2-a-b\}+2ab\{c-1\}=0\\ But\ \ a,\ b,\ c\ \neq\ 0.\\ So\ \ c-1=0\ \ \implies\ \color{#3D99F6}{c=1}.\\ and\ \ a^2+b^2-a-b=0.\\ \implies\ a^2+b^2= a+b.\\ This\ is\ only\ possible\ for\ integer\ values\ if\ \ a=a^2\ or\ b^2, \ with\ b= b^2\ or\ a^2.\\ Only \ such\ values\ are\ a=1\ and\ b=1.\\ \text{So ABC can take only ONE value.}$

Let $AB=x$ , $AC=y$ , and $BC=z$ .

This will turn $[ABHJ]=[BCDE]+[ACFG]$ into $xy(y-1)+xz(z-1)+2yz(x-1)=0$ .

Because $AB$ , $AC$ , and $BC$ are of integer length, $x$ , $y$ , and $z$ are all integers equal or greater than $1$ . But then $xy(y-1) \geq 0$ (equality in the case of $y=1)$ , $xz(z-1) \geq 0$ (equality in the case of $z=1)$ , and $2yz(x-1) \geq 0$ as well (equality in the case of $x=1)$ .

So the only possible solution to $xy(y-1)+xz(z-1)+2yz(x-1)=0$ is for all terms on the LHS to equal $0$ , resulting in $(x,y,z)=(1,1,1)$ .

Therefore the number of possible values of $[ABC]$ is $\fbox{1}$ .