Try this beautiful Problem on Geometry based on Circle and Equilateral Triangle from AMC 10 A, 2017. You may use sequential hints to solve the problem.
Sides $\overline{A B}$ and $\overline{A C}$ of equilateral triangle $A B C$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle A B C$ lies outside the circle? $?$
,
Geometry
Triangle
Circle
Pre College Mathematics
AMC-10A, 2017 Problem-22
$\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$
Given that ABC is a equilateral triangle whose \(AB\) & \(AC\) are the tangents of the circle whose centre is \(O\). We have to find out the fraction of the area of $\triangle A B C$ lies outside the circle
we have to find out thr ratio of the areas of Blue colour : Red colour area. Therefore we have to findout the area of the circle and Triangle ABC.
Later we have to find out red area and subtract from the Triangle ABC.
Now can you finish the problem?
Let the radius of the circle be $r$, and let its center be $O$. since $\overline{A B}$ and $\overline{A C}$ are tangent to circle $O$, then $\angle O B A=\angle O C A=90^{\circ}$, so $\angle B O C=120^{\circ} .$ Therefore, since $\overline{O B}$ and $\overline{O C}$ are equal to $r$, then $\overline{B C}=r \sqrt{3}$. The area of the equilateral triangle is $\frac{(r \sqrt{3})^{2} \sqrt{3}}{4}=\frac{3 r^{2} \sqrt{3}}{4},$ and the area of the sector we are subtracting from it is $\frac{1}{3} \pi r^{2}-\frac{1}{2} r \cdot r \cdot \frac{\sqrt{3}}{2}=\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4} .$
Now Can you finish the Problem?
Therefore the area outside the circle is $\frac{3 r^{2} \sqrt{3}}{4}-\left(\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4}\right)=r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}$
Therefore the Required fraction is $\frac{r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}}{\frac{3 r^{2} \sqrt{3}}{4}}$
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Try this beautiful Problem on Geometry based on Circle and Equilateral Triangle from AMC 10 A, 2017. You may use sequential hints to solve the problem.
Sides $\overline{A B}$ and $\overline{A C}$ of equilateral triangle $A B C$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle A B C$ lies outside the circle? $?$
,
Geometry
Triangle
Circle
Pre College Mathematics
AMC-10A, 2017 Problem-22
$\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$
Given that ABC is a equilateral triangle whose \(AB\) & \(AC\) are the tangents of the circle whose centre is \(O\). We have to find out the fraction of the area of $\triangle A B C$ lies outside the circle
we have to find out thr ratio of the areas of Blue colour : Red colour area. Therefore we have to findout the area of the circle and Triangle ABC.
Later we have to find out red area and subtract from the Triangle ABC.
Now can you finish the problem?
Let the radius of the circle be $r$, and let its center be $O$. since $\overline{A B}$ and $\overline{A C}$ are tangent to circle $O$, then $\angle O B A=\angle O C A=90^{\circ}$, so $\angle B O C=120^{\circ} .$ Therefore, since $\overline{O B}$ and $\overline{O C}$ are equal to $r$, then $\overline{B C}=r \sqrt{3}$. The area of the equilateral triangle is $\frac{(r \sqrt{3})^{2} \sqrt{3}}{4}=\frac{3 r^{2} \sqrt{3}}{4},$ and the area of the sector we are subtracting from it is $\frac{1}{3} \pi r^{2}-\frac{1}{2} r \cdot r \cdot \frac{\sqrt{3}}{2}=\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4} .$
Now Can you finish the Problem?
Therefore the area outside the circle is $\frac{3 r^{2} \sqrt{3}}{4}-\left(\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4}\right)=r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}$
Therefore the Required fraction is $\frac{r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}}{\frac{3 r^{2} \sqrt{3}}{4}}$
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
