Categories

## Ratio of Circles | AMC-10A, 2009 | Problem 21

Try this beautiful problem from Geometry based on ratio of Circles from AMC 10A, 2009, Problem 21.

## Ratio of Circles – AMC-10A, 2009- Problem 21

Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?

• $3-2 \sqrt{2}$
• $2-\sqrt{2}$
• $4(3-2 \sqrt{2})$
• $\frac{1}{2}(3-\sqrt{2})$
• $2 \sqrt{2}-2$

### Key Concepts

Geometry

Circle

Pythagoras

Answer: $4(3-2 \sqrt{2})$

AMC-10A (2009) Problem 21

Pre College Mathematics

## Try with Hints

We have to find out the ratio of the sum of the areas of the four smaller circles to the area of the larger circle. To find out the area any circle,we need radius.so at first we have to find out radius of two types circle.Can you find out the radius of two type circle i.e small circle and big circle………………

Can you now finish the problem ……….

Let the radius of the Four small circles be $r$.Therfore from the above diagram we can say $CD=DE=EF=CF=2r$. Now the quadrilateral $CDEF$ in the center must be a square. Therefore from Pythagoras theorm we can say $DF=\sqrt{(2r)^2 + (2r)^2}=2r\sqrt 2$. So $AB=AD+DF+BF=r+2r\sqrt 2+r=2r+2r\sqrt 2$

Therefore radius of the small circle is $r$ and big circle is$R=r+r \sqrt{2}=r(1+\sqrt{2})$

Can you now finish the problem ……….

Therefore the area of the large circle is $L=\pi R^{2}=\pi r^{2}(1+\sqrt{2})^{2}=\pi r^{2}(3+2 \sqrt{2})$and the The area of four small circles is $S=4 \pi r^{2}$

The ratio of the area will be $\frac{S}{L}=\frac{4 \pi r^{2}}{\pi r^{2}(3+2 \sqrt{2})}$

=$\frac{4}{3+2 \sqrt{2}}$

=$\frac{4}{3+2 \sqrt{2}} \cdot \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}$

=$\frac{4(3-2 \sqrt{2})}{3^{2}-(2 \sqrt{2})^{2}}$

=$\frac{4(3-2 \sqrt{2})}{1}$

=$4(3-2 \sqrt{2})$

Categories

## Area of the Inner Square | AMC-10A, 2005 | Problem 8

Try this beautiful problem from Geometry: Area of the inner square

## Area of Inner Square – AMC-10A, 2005- Problem 8

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$. What is the area of the inner square $EFGH$?

• $25$
• $32$
• $36$
• $42$
• $40$

### Key Concepts

Geometry

Square

similarity

Answer: $36$

AMC-10A (2005) Problem 8

Pre College Mathematics

## Try with Hints

We have to find out the area of the region $EFGH$ Which is a square shape .so if we can find out one of it’s side length then we can easily find out the area of $EFGH$. Now given that $BE=1$ i.e $BE=CF=DG=AH=1$ and side length of the square $ABCD=\sqrt {50}$.Therefore $(AB)^2=(\sqrt {50})^2=50$.so using this information can you find out the length of $EH$?

Can you find out the required area…..?

Since $EFGH$ is a square,therefore $ABH$ is a Right -angle Triangle.

Therefore,$(AH)^2+(BH)^2=(AB)^2$

$\Rightarrow (AH)^2+(HE+EB)^2=(AB)^2$

$\Rightarrow (1)^2+(HE+1)^2=50$

$\Rightarrow (HE+1)^2=49$

$\Rightarrow (HE+1)=7$

$\Rightarrow HE=6$

Therefore area of the inner square (red shaded region) =${6}^2=36$

Categories

## Right angled triangle | AIME I, 1994 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Right angled triangle.

## Right angled triangle – AIME I, 1994

In $\Delta ABC$, $\angle C$ is a right angle and the altitude from C meets AB at D. The lengths of the sides of $\Delta ABC$ are integers, $BD={29}^{3}$, and $cosB=\frac{m}{n}$, where m, n are relatively prime positive integers, find m+n.

• is 107
• is 450
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Right angled triangle

Pythagoras Theorem

AIME I, 1994, Question 10

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

$\Delta ABC \sim \Delta CBD$

$\frac{BC}{AB}=\frac{29^{3}}{BC}$

$\Rightarrow {BC}^{2}=29^{3}AB$

$\Rightarrow 29^{2}|BC and 29|AB$

$\Rightarrow BC and AB are in form 29^{2}x, 29x^{2}$ where x is integer

$by Pythagoras Theorem, AC^{2}+BC^{2}=AB^{2}$

$\Rightarrow (29^{2}x)^{2}+AC^{2}=(29x^{2})^{2}$

$\Rightarrow 29x|AC$

taking y=$\frac{AC}{29x}$ and dividing by (29x)^{2}\)

$\Rightarrow 29^{2}=x^{2}-y^{2}=(x-y)(x+y)$

where x,y are integers, the factors are $(1,29^{2}),(29,29)$

$y=\frac{AC}{29x}$ not equals 0 $\Rightarrow x-y=1, x+y=29$

$\Rightarrow x=\frac{1+29^{2}}{2}$

=421 then$cosB=\frac{BC}{AB}=\frac{29^{2}x}{29x^{2}}$=$\frac{29}{421}$

m+n=29+421=450.