A geometry problem by Ajay Sambhriya

Proof for Pythagorean Theorem springs to mind.

We can form a larger square of sidelength $\overline{AB}+\overline{BC}$ , with one corner at $B$ . The diagonal length $\overline{BG}=20$ , so the area of the large square $FGHB=\frac{20\times 20}{2}=200$ .

The area of quadrilateral $ABCO=\frac{200}{4}=50$ .

Using the Distance Formula, we can write an equation for the line segments AB and BC.

$\displaystyle\overline{AB}=\sqrt{|y-1|^2+|x-0|^2}$ $\Rightarrow\displaystyle\overline{AB}=\sqrt{x^2+y^2-2y+1}$

$\displaystyle\overline{BC}=\sqrt{|y-1|^2+|x-1|^2}$ $\Rightarrow\displaystyle\overline{BC}=\sqrt{x^2+y^2-2x-2y+2}$

It is important to note that the lines AB and BC are perpendicular to each other since they form a right angle.

And we know that the product of the slopes of perpendicular lines is -1.

$\displaystyle m\overline{AB}=\frac{y-1}{x}$

$\displaystyle m\overline{BC}=\frac{y-1}{x-1}$

$\displaystyle m\overline{AB}\cdot m\overline{BC}=-1$

$\boxed{\displaystyle x^2+y^2=x+2y-1}$

Let's use this equation to simplify the $\displaystyle\overline{AB}$ and $\displaystyle\overline{BC}$ .

$\displaystyle\overline{AB}=\sqrt{(x^2+y^2)-2y+1}$

$\displaystyle\overline{AB}=\sqrt{(x+2y-1)-2y+1}$

$\displaystyle\overline{AB}=\sqrt{x}$

From the figure above, we know the equation: $\displaystyle x^2+y^2=\overline{AB}^2$

Substituting..

$\displaystyle(x^2+y^2)=\overline{AB}^2$

$\displaystyle(x+2y-1)=(\sqrt{x})^2$

$\displaystyle x+2y-1=x$

$\displaystyle y=\frac{1}{2}$

$\displaystyle x=\frac{1}{2}$

Hence the ordered pair of point B is $\displaystyle\left(\frac{1}{2},\frac{3}{2}\right)$ .

So, to write a more specific drawing, we have this:

The line $\displaystyle\overline{BO}$ is a vertical straight line.

Since we used a unit circle in representing the problem in Cartesian Plane, the length $\displaystyle\overline{BO}$ has a factor.

The length of $\displaystyle\overline{BM}$ and $\displaystyle\overline{MO}$ adds to a total of 1.

And according to the problem, $\displaystyle\overline{BO}$ = 10

Therefore, there is a factor of 10. Let's call this $\displaystyle k=10$

Also, if $\displaystyle\overline{AC}$ =1, then its actual measurement is $\displaystyle k$ times $\displaystyle\overline{AC}$ .

$\displaystyle k\cdot$ $\displaystyle\overline{AC} = 10$

$\displaystyle\overline{BO}$ and $\displaystyle\overline{AC}$ are the diagonals of the square $ABCO$ .

Area of the square using the diagonals $= \displaystyle\frac{d_1d_2}{2}$

Area $= \displaystyle\frac{(10)(10)}{2}$

Area $\displaystyle= 50 cm^2$

Let's put a pont F at the middle of line AC. Let AF=FC=r. Note that r is a radius of a circle that has ABCO inscribed. Using triangle BOF and law of cosines we write 10^2=2r^2(1+sin2A) where A is the angle BAC. This is because angle BFO=pi/2+2A. The area of ABCO is P=r^2+(2r sinA)(2r cosA)/2=r^2(1+sin2A) which is 100/2=50. We used that area is a sum of areas of triangles ABC and ACO. In general, area is always (OB^2)/2.

We can also use the law of cosine and sine to find the solution:

(1) $AB^{2}$ = $AO^{2}$ + $BO^{2}$ - 2 * AO * BO * cosAOB

(2) $BC^{2}$ = $BO^{2}$ + $CO^{2}$ - 2 * BO * CO * cosBOC

(3) AOC is a right triangle and AO = CO = $\frac{\sqrt{2}}{2}$ AC

(4) We also know triangle ABC is a right triangle and $AB^{2}$ + $BC^{2}$ = $AC^{2}$ , from (1), (2) and (3) we can derive the following equation:

$AC^{2}$ = $\frac{1}{2}$ $AC^{2}$ + 100 - 10 $\sqrt{2}$ * AC * CosAOB + $\frac{1}{2}$ $AC^{2}$ + 100 - 10 $\sqrt{2}$ * AC * CosBOC

=> $\sqrt{2}$ * AC * (CosAOB+CosBOC) = 20

(5) According to the law of sine, the area of triangle AOB = $\frac{1}{2}$ * AO * BO * SinAOB and the area of triangle BOC = $\frac{1}{2}$ * BO * CO * SinBOC

(6) Since AOC is a right triangle, then CosAOB = SinBOC and CosBOC = SinAOB => CosAOB+CosBOC=SinAOB+SinBOC

(7) Therefore, area of ABCO = $\frac{1}{2}$ * $\frac{\sqrt{2}}{2}$ AC * 10 * (CosAOB+CosBOC) = 2.5 * $\sqrt{2}$ * AC * (CosAOB+CosBOC)

(8) Finally, from (4), the area of ABCO = 2.5 * 20 = $\boxed{50}$

if BO = 2

At first glance, there isn't enough information to solve the problem, because angle ABO cannot be determined. Given that a unique solution exists without this information, I am free to draw the figure any way I please. I choose BO as the bisector of AOC. Given that ABC is a right angle, then ABCO is just a square with diagonal 10. Pythagoras gives the side of that square as 5sqrt(2), and area 50. Note that we weren't asked to prove that this answer holds for all angles ABO

Consider BC and BA as a part of diagonals of square say ACFG drawn aside ACDE.

Let AC=x Therefore FC=x AD is diagonal. AD=(root 2)x AO=OC=x/(root 2)
(AO=AD/2) Similarly. BC=BA=x/(root 2)

Therefore.ABCO is square

Area=(x square)/2

Now. From B and O draw perpendiculars BM and ON on FC and CD respectively

Therefore.

MC=CN=x/2 MN=MC+CN=x

But.MN=OB

x=10 Therefore. Area of square ABCO=50. (Ans)

Let AB be y, BC be y, AO and AC be r.

By Pythagoras' Theorem , x² + y² = 2r², and AC = r√2

Then by Ptolemy Theorem , xr + yr = 10(√2)(r) ⇒ x + y = 10√2

By squaring both side we get x² + y² + 2xy = 200, sub x² + y² = 2r² into the equation and get r² + xy = 100

Since it ask for area, area of ABCO = (r² + xy)/2 = 100/2 = 50

Since angle ABC is given as 90 degrees, we know that side AC is the diameter of a semicircle which circumscribes triangle ABC. Since ACDE is a square, symmetry conditions demand that all 4 angles around O are identical, 90 degrees also. The altitude of triangle AOC passes through P, the bisector of segment AC, and is by definition perpendicular to AC. As O is the center of the square, symmetry requires that the altitude just drawn have a length of 5. Since P is the center of the circumscribed circle (for both right triangles), the distance to points A, B, C, and O is the same for all four, 5. Since OP + PB = 10, the Law of Cosines requires that the three points must be collinear, and the two triangles are congruent, with a total area equal to 1/2 of the square, ACDE, 50.

Angle AOC = 90° And side AD = side AC (diagonals of square bisect each other at 90°)

And given that angle ABC = 90° Which implies ABCO is a square whose diagonal is OB .

Let a be the side of the square ABCO , so OB = $a\sqrt{2}$

Or a = OB / $\sqrt{2}$ = $\frac {10}{\sqrt{2}}$

Now Area(ABCO) = $a^2$ = 50

$consider AB=x , BC=y \text{ since B+O=180 so A,B,C,O are cyclic . we can use ptolemy's equality }\\ \\$ $AC.BO=CO.x +AO.y \\ or , 10AC=(x+y)AO [\text{ since } AO=CO=\frac{1}{\sqrt2}AC] \\ or, x+y=10\sqrt2 \\ or(x+y)^2=200 \\ or, x^2+y^2+2xy=200 \\ or, AC^2+2xy=200 \text{ [ since } x^2+y^2=AC^2 ]\\ or, AC . \frac{AC}{2} .2 +2xy=200\\ or,AC.h.2+2xy=200 [ \text { if AC is considered as base the height of }\Delta AOC = h=\frac{AC}{2} ]\\ or,4(\frac{1}{2}AC.h+\frac{1}{2}.xy)=200\\ or, 4(\Delta AOC+\Delta ABC)=200 \\ or, 4ABCO=200 \\ or, ABCO=50 \text{ which is the area :) }$

consider the direction of axes along AC and CD with origin at C, and the problem simplifies as followed:-