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AMC 10 Geometry Math Olympiad USA Math Olympiad

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?

Figure of the problem
  • $3-2 \sqrt{2}$
  • $2-\sqrt{2}$
  • $4(3-2 \sqrt{2})$
  • $\frac{1}{2}(3-\sqrt{2})$
  • $2 \sqrt{2}-2$

Key Concepts


Geometry

Circle

Pythagoras

Check the Answer


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

AMC-10A (2009) Problem 21

Pre College Mathematics

Try with Hints


Circles in Circle

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 ……….

finding ratio of area of circles

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 ……….

shaded circles

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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AMC 10 Math Olympiad USA Math Olympiad

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 $EFGH$?

Area of the Inner Square - Problem Figure
  • \(25\)
  • \(32\)
  • \(36\)
  • \(42\)
  • \(40\)

Key Concepts


Geometry

Square

similarity

Check the Answer


Answer: \(36\)

AMC-10A (2005) Problem 8

Pre College Mathematics

Try with Hints


Area of the Inner Square - Shaded Figure

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…..?

Explanatory Shading of the figure

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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Algebra Arithmetic Geometry Math Olympiad PRMO

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

Check the Answer


Answer: is 48.

PRMO, 2019, Question 25

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

Let radius =r

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

\(=\sqrt{256+32r}\)

or, \(AD^{2}+DC^{2}=CA^{2}\)

Diameter of a circle - figure

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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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 450.

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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Singapore Math Olympiad

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

Check the Answer


Answer: 65

Singapore Mathematical Olympiad

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 .

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