Triangle $ABC$ has integer side lengths. Rectangles $BCDE, ACFG, ABHJ$ are constructed so that $CD = AC + AB$ , $CF = AB + BC$ , and $BH = (AC + BC)^2$ . If $[ABHJ] = [BCDE] + [ACFG]$ , how many different values can $[ABC]$ have?

**
Details and assumptions
**

$[PQRS]$ refers to the area of figure $PQRS$ .

The answer is 1.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

10 Helpful
0 Interesting
0 Brilliant
0 Confused

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 $1$

10 Helpful
0 Interesting
0 Brilliant
0 Confused

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

10 Helpful
0 Interesting
0 Brilliant
0 Confused

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.

10 Helpful
0 Interesting
0 Brilliant
0 Confused

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
$[ABC]$
.

10 Helpful
0 Interesting
0 Brilliant
0 Confused

$[ABHJ] = [BCDE] + [ACFG]$ 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.

10 Helpful
0 Interesting
0 Brilliant
0 Confused

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

5 Helpful
0 Interesting
0 Brilliant
0 Confused

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

5 Helpful
0 Interesting
0 Brilliant
0 Confused

5 Helpful
0 Interesting
0 Brilliant
0 Confused

2 Helpful
0 Interesting
0 Brilliant
0 Confused

0 Helpful
0 Interesting
0 Brilliant
0 Confused

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 $\fbox{1}$ .

0 Helpful
0 Interesting
0 Brilliant
0 Confused

×

Problem Loading...

Note Loading...

Set Loading...

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 $\boxed {1}$ .