Sides Extended !!!

title

Reference image 1

$a:y=0\\ b:x=0\\ c:y=1\\ d:x=1\\ e:y=-2x+1\\ f:y=\frac { x }{ 2 } \\ A(0,1)\\ B(0,0)\\ C(1,0)\\ D(1,1)\\ E(2,1)\\ F(1,-1)\\ O(x,y)$

I - Discovering O's coordinates:

$\frac { x }{ 2 } =1-2x\\ \frac { x }{ 2 } +2x=1\\ \frac { 5x }{ 2 } =1\\ x=\frac { 2 }{ 5 } =0.4\\ y=\frac { 0.4 }{ 2 } =-2\times (0.4)+1=0.2\\ O(0.4,0.2)$

II - Discovering $\frac { OE }{ EB }$

Reference Image 2

$P1(0.4,0)\\ P2(2,0.2)\\ \angle OBP1\quad and\quad \angle EOP2\quad are\quad congruent\\ \angle BOP1\quad and\quad \angle OEP2\quad are\quad congruent\\ \triangle BOP1\quad and\quad \triangle OEP2\quad are\quad similar$

$\frac { OE }{ BO } =\frac { OP2 }{ BP1 } =\frac { P2E }{ P1O } =\frac { m }{ n }$

$\frac { OP2 }{ BP1 } =\frac { 2-0.4 }{ 0.4-0 } =\frac { 1.6 }{ 0.4 } =\frac { 4 }{ 1 } =\frac { m }{ n }$

$m=4; n=1$

III - Summing m and n $m+n=4+1=5$

So, the answer is 5

Notice that triangle AOB and triangle EOA and triangle EAB are similar. Since AB/AE = 1/2 = OB/OA = OA/OE. So OE = 4*OB so the required fraction is 4/1 so the answer is 5.

$\frac{OE}{AE} = \frac{DF}{AF}$

$\frac{OE}{2} = \frac{2}{\sqrt5}$

$OE = \frac{4\sqrt5}{5}$

$OB = BE - OE$

$OB = \sqrt5 - \frac{4\sqrt5}{5} = \frac{\sqrt5}{5}$

$\frac{OE}{OB} = \frac{4}{1}$

$m + n = 4 + 1 =\boxed{5}$

Let $\angle BAF = \angle AFD = \angle x$ and $\angle AOB = y$

In $\triangle AOB$ ,

$\frac{\sin BAO}{OB} = \frac{\sin AOB}{AB}$

$\Rightarrow \frac{\sin x}{OB} = \frac{\sin y}{AB}$ _ _ _ (1)

$$

In $\triangle AOE$ ,

$\frac{\sin OAE}{OE} = \frac{\sin AOE}{2AD}$ $[AD = DE]$

$\Rightarrow \frac{\sin (90-x)}{OE} = \frac{\sin (180-y)}{2AD}$

$\Rightarrow \frac{\cos x}{OE} = \frac{\sin y}{2AD}$ _ _ _ (2)

$$

Again in $\triangle ADF$ ,

$\tan x = \frac{AD}{2AB}$ $[AB=CD]$

$$

Divide $(1)$ by $(2)$ .....

$\tan x \frac{OE}{OB} = \frac{2AD}{AB}$

$\frac{OE}{OB} = \frac{2AD}{\tan x \times AB}$

$\frac{OE}{OB} = \frac{2AD \times 2AB}{AD \times AB}$

$\frac{OE}{OB} = 4 = \frac{4}{1} = \frac{m}{n}$

Hence, $m+n = 4+1 = \boxed{5}$

Ruan's answer is similar to mine, but takes a very long workaround once O is found. Here is a shorter version:

Assume the lower left corner of the square is the origin of the Cartesian x-y plane; in other words, it is (0,0).

• B: (0,0)
• A: (0, 1)
• E: (2, 1)
• F: (1, -1)

From these points, it is relatively easy to deduce that the line AF is $y = -2x + 1$ , and the line BE is $y = \frac{x}{2}$ . These lines intersect at $-2x+1 = \frac{x}{2}$ $\frac{5x}{2} = 1$ $x = \frac{2}{5}$ Plugging this x value into line BE returns $y - \frac{1}{5}$

So now we have point O: (\frac{2}[5}, \frac{1}{5}) . Since AEB is a triangle, we can just use the ratio of the y-differences from B to O and from O to E. Thus, $\frac{OE}{OB} = \cfrac{\frac{4}{5}}{\frac{1}{5}}$ $\frac{OE}{OB} = \frac{1}{4}$ And therefore, $m+n = \boxed{5}$

It can be seen that $\Delta AOB$ and $\Delta AOE$ are similar to $\Delta ABE$ .

Therefore, $\frac{OE}{OA} = \frac{AE}{AB} = \frac{2}{1}$ .

Similarly, $\frac{OA}{OB} = \frac{AE}{AB} = \frac{2}{1}$ .

Now, $\frac{OE}{OB} = \frac{OE}{OA} \times \frac{OA}{OB} = \frac{2}{1} \times \frac{2}{1} = \frac {4}{1}$ .

Therefore, $m=4$ and $n=1$ ; $m+n=\boxed{5}$ .

Just look at diagram and its easy :v

Let BE intersect DE at G, AF intersect BC at N. Let sides of square be s.
Draw GH parallel to BC to intersect AB at G and AF at K.
From triangle AFD, NC = s/2. So BN = s/2.
In triangle ABN, H is midpoint.
We can see GK =(1/2)BN = s/4. So KG = s - s/4 =3s/4
IN similar triangles KGO and NBO...>>. GO/OB =KG/BN = (3s/4)/(s/2)
= (3/2) * BG = GE = BO + GO = (5/2)BO
So...>> OE/BO = (OG + GE)/BO = {..(3/2) * BO..+..(5/2) * BO..}/BO = 8/2 = 4/1 = a/b... a+b = 5

1- let side of square be a. therefore AB = a

2- let AF intersect BC at P

3- notice that OB is the altitude for AOB

4- hence area of AOB = [ .5 * AP *BO ] = [ .5 * AB * BP]

5- AP = a * root5/2 (pythagoras on ABP)......AB = a (side of square)....BP = a/2 (half of side)

6- therefore BO = n = a/root5

7- BE =a* root5

8- therefore m = 4a/root5

9 - therefore m+n = 5