Right-Angle Point

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There is probably a more elegant way of doing this, but I solved this using the Law of Cosines first on $\Delta AOB$ , then again on $\Delta AOC$ .

Here are the steps:

• Applying the Law of Cosines on $\Delta AOB$ with respect to $θ_{1}$ : $(5\sqrt{5})^{2}=(5\sqrt{5})^{2}+10^{2}-2(10)(5\sqrt{5})\cos{θ_{1}} \Rightarrow \cos{θ_{1}}=\frac{1}{\sqrt{5}}$ .
• We can also observe, since $∠ACB=90^{o}$ , that $\tan∠CAB=\frac{3}{4}$ .
• Since $m∠CAO=m∠CAB-θ_{1}$ , we can find $\cos∠CAO$ : $\cos∠CAO=\cos∠CAB \cosθ_{1}+\sin∠CAB \sinθ_{1}=(\frac{4}{5})(\frac{1}{\sqrt{5}})+(\frac{3}{5})(\frac{2}{\sqrt{5}})=\frac{2}{\sqrt{5}}$ .
• Applying the Law of Cosines on $\Delta AOC$ : $(OC)^{2}=8^{2}+(5\sqrt{5})^{2}-2(8)(5\sqrt{5})cos∠CAO=189-16(5\sqrt{5})(\frac{2}{\sqrt{5}})=\fbox{29}$ .

Let the altitude from C to AB be E, the altitude from O to AB be D, and let us put the entire diagram on a coordinate plane, with A = (0,0) and all defined points in the first quadrant (or the boundaries of the first quadrant.)

Note that CE, EA, and AC form a right triangle. As the area of ABC is 24, and the length of AC is 10, we must have that the altitude to AC, or CE, must be 4.8. By the Pythagorean theorem on ACE, it is also an elementary result that AE = 6.4.

Now, note that OAB is an isosceles triangle with leg lengths $5\sqrt{5}$ and base $10$ . Let the altitude from O to AB be D. By the Pythagorean theorem, we have that $AD=5$ . Thus, O = (5,10)

Thus we have C = (4.8,6.4) and that O = (5,10). Using the distance formula, we find that the square of the distance between these points is 29.

Let the midpoint of $AB$ be $M$ . Note that by Pythagorean Theorem, $OM^2=MA^2+AO^2$ and plugging values in and solving yields $OM=10$ . Now let the altitude of $\Delta ABC$ from $C$ hit $AB$ at $D$ . Note that $DM=\dfrac{7}{5}$ and $CD=\dfrac{26}{5}$ from similar triangles. Using Pythagorean Theorem again, we get that $OC=\sqrt{29}$ . Therefore, $OC^2=\boxed{29}$ .

$$ Draw a circle in Cartesian diagram where its center at the origin and $\Delta\text{ABC}$ in the first quadrant. Let $\,(a,b)$ be the coordinate of point $\,C$ , then it's obvious the coordinates of $\,A$ and $\,B$ are $\,(a,b+6)$ and $\,(a+8,b)$ , or vice versa. The equation of the circle is $$ $\begin{aligned} x^2+y^2&=r^2\\ x^2+y^2&=125, \end{aligned}$ $$ then substitute the coordinates of $\,A$ and $\,B$ into the equation. We obtain $$ $\begin{aligned} (a+8)^2+b^2=125 & & \text{and}& & a^2+(b+6)^2=125. \end{aligned}$ $$ The solutions of those equations are: $\;a_1=-10$ ; $\;b_1=-11$ and $\;a_2=2$ ; $\;b_2=5$ . Since $\Delta\text{ABC}$ in the first quadrant, we use $\;a_2=2$ and $\;b_2=5$ . Thus, $\;OC^2=a_2^2 + b_2^2 = \boxed{29}$ . $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Let us first consider the triangle OAB., since it is isosceles, the height can easily be found from O. It is 10.

Thus, area = 50 sq. units. on equating this with $\frac{1}{2}5\sqrt{5}^{2}sinA$ , we get sinA=4/5, that means A=53 degrees. Hence, angle OAB=OBA=63.5

It is clear that the triangle ABC is inside OBC (since its height is 4.8 which is less than 10) and its angles are 53,37 and 90.

Hence, angle OAC=63.5-37=26.5 (53/2).

We also know that $\cos \theta = \sqrt{\frac{\cos \theta + 1}{2}}$

Hence, applying cosine formula in OCA, we get $\cos \frac{53}{2} = \frac{125 + 64 - x^{2}}{80\sqrt{5}}$ which gives the value of x(OC) as $\sqrt{29}$ . Hence $\boxed{OC^{2}= 29}$

Since OA>CA, OB>CB then the point C is inside the triangle OAB. cosOAB = (OA^2 + AB^2 - OB^2)/ (2 x OA x AB) cosCAB = AC/AB = 8/10 angle OAC = angle OAB - angle CAB OC^2 = OA^2 + AC^2 - 2 x AC x OA x cosOAC = 29

Primeiro passo é transformar o problema em Geometria Analítica. Assim, vamos supor que: O = (0,0) ; B=(x,y) ; A=(x-6 ,y+8) ; C=(x-6, y). A distância de O para A e B, é 5\sqrt{5}. Calculando a distância, ficará:

$125 = X^2+Y^2$

$125 = X²+Y²-12x+36+16y+64$

Então: $-12x+36+16y+64 = 0 \rightarrow (4y+25)/3 = x$ Substituindo : 125*9 = 16y²+200y+625+9y² = > 1125 = 25y²+200y+625 = >0 = y² + 8y -20

O valor positivo de y é 2, Portanto y=2. Assim: x=11 C = (5, 2)

Calculando a distância entre o ponto O ao ponto C, vai ser 5²+2² = OC² = > 29 = OC²

$\angle BAO= \arccos \left(\frac{100+125-125}{100 \sqrt{5}} \right)$

$\angle BAC=\arccos \left( \frac{4}{5} \right)$

$\cos( \angle BAO-\angle BAC)=x=\frac{2}{\sqrt{5}}$

$OC^2=125+64-2 \times \frac {2}{\sqrt{5}} \times 5\sqrt{5} \times 8 =\fbox {29}$