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## 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 $\frac{S}{L}=\frac{4 \pi r^{2}}{\pi r^{2}(3+2 \sqrt{2})}$

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

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

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

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

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

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## 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) =${6}^2=36$

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## Diameter of a circle | PRMO 2019 | Question 25

Try this beautiful problem from the Pre-RMO, 2019 based on Diameter of a circle.

## Diameter of a circle – PRMO 2019

A village has a circular wall around it, and the wall has four gates pointing north, southeast and west. A tree stands outside the village, 16 m north of the north gate, and it can be just seen appearing on the horizon from a point 48 m east of the south gate. Find the diameter in meters of the wall that surrounds the village.

• is 107
• is 48
• is 840
• cannot be determined from the given information

### Key Concepts

Pythagoras Theorem

Equations

Integer

PRMO, 2019, Question 25

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

or,$AB=\sqrt{AO^{2}-OB^{2}}=\sqrt{(16+r)^{2}-r^{2}}$

$=\sqrt{256+32r}$

or, $AD^{2}+DC^{2}=CA^{2}$

Second Hint

or, $48^{2}+(2r+16)^{2}$

$=(48+\sqrt{256+32r})^{2}$

or, $r^{2}+8r=24\sqrt{256+32r}$

or, $r(r+8)=24(4\sqrt{2})(\sqrt{r+8})$

Final Step

or, r=24

or, 2r=diameter=48.

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## 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

First hint

$\Delta ABC \sim \Delta CBD$

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

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

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

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

Second Hint

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

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

$\Rightarrow 29x|AC$

Final Step

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.

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## Application of Pythagoras Theorem | SMO, 2010 | Problem 22

Try this problem from the Singapore Mathematics Olympiad, SMO, 2010 based on the application of the Pythagoras Theorem.

## Application of Pythagoras Theorem- (SMO Test)

The figure below shows a circle with diameter AB. C ad D are points on the circle on the same side of AB such that BD bisects $\angle {CBA}$. The chords AC and BD intersect at E. It is given that AE = 169 cm and EC = 119 cm. If ED = x cm, find the value of x.

• 65
• 55
• 56
• 60

### Key Concepts

Circle

Pythagoras Theorem

2D – Geometry

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

If you get stuck in this problem this is the first hint we can start with :

As BE intersect $\angle {CBA}$ we have $\frac {BC}{BA} = \frac {EC}{EA} = \frac {119}{169}$

Thus we can let BC = 119 y and BA = 169 y .

Since $\angle {BCA} = 90 ^\circ$.

Then try to do the rest of the problem ………………………………………………

If we want to continue from the last hint we have :

Apply Pythagoras Theorem ,

$AB ^2 = AC^2 + BC ^2$

$(169y)^2 = (169 + 119)^2 + (119y)^2$

$y^2 (169-119)(169+119) = (169+119)^2$

$y^2 = \frac {169+119}{169-119} = \frac {144}{25}$

$y = \frac {12}{5}$

In the last hint:

Hence , from triangle BCE , we have BE = $\sqrt{119^2 + (119y)^2} = 119 \times \frac {13}{5}$

Finally , note that $\triangle {ADE}$ and $\triangle {BCE}$ are similar , so we have

ED = $\frac {AE \times CE}{BE} = \frac {169 \times 119}{119 \times \frac {13}{5}} = 65$ cm .