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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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Algebra Arithmetic Geometry Math Olympiad PRMO

Triangle and Integer | PRMO 2019 | Question 28

Try this beautiful problem from the Pre-RMO, 2019 based on Circles,Triangle and largest integer.

Circles, Triangle and largest Integer – PRMO 2019


Let ABC be a triangle with sides 51, 52, 53. Let D denote the incircle of triangle ABC. Draw tangents to D which are parallel to the sides ABC. let \(r_1\). \(r_2\), \(r_3\) be the inradii of the three corner triangles so formed, find the largest integer that does not exceed \(r_1+r_2+r_3\).

Triangle and Integer
  • is 107
  • is 15
  • is 840
  • cannot be determined from the given information

Key Concepts


Circles

Triangle

Largest Integers

Check the Answer


Answer: is 15.

PRMO, 2019, Question 28

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

Let PQ be one of tangents parallel to BC and meet sides AB and AC at P and Q let PQ=x and BC=51

Second Hint

triangle ABC similar with triangle APQ then\(\frac{x}{a}=\frac{r_1}{r}=\frac{s-a}{s}\) which is in same way for \(\frac{y}{b}\) and \(\frac{z}{c}\) then \(\frac{r_1+r_2+r_3}{r}\)=\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\)=3-2=1

Final Step

then \(r_1+r_2+r_3\)=r and r by given condition of question =\((\frac{s(s-a)(s-b)(s-c)}{s})^\frac{1}{2}\)=\((\frac{78(78-51)(78-52)(78-53)}{78})^\frac{1}{2}\)=15.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Cones and circle | AIME I, 2008 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Cones and circle.

Cones and circle – AIME I, 2008


A right circular cone has base radius r and height h the cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cones base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making 17 complete rotations. The value of \(\frac{h}{r}\) can be written in the form \(m{n}^\frac{1}{2}\) where m and n are positive integers and n in not divisible by the square of any prime, find m+n.

  • is 107
  • is 14
  • is 840
  • cannot be determined from the given information

Key Concepts


Cones

Circles

Algebra

Check the Answer


Answer: is 14.

AIME I, 2008, Question 5

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

The path is circle with radius =\(({r}^{2}+{h}^{2})^\frac{1}{2}\) then length of path=\(2\frac{22}{7}({r}^{2}+{h}^{2})^\frac{1}{2}\)

Second Hint

length of path=17 times circumference of base then \(({r}^{2}+{h}^{2})^\frac{1}{2}\)=17r then \({h}^{2}=288{r}^{2}\)

Final Step

then \(\frac{h}{r}=12{(2)}^\frac{1}{2}\) then 12+2=14.

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Geometry Math Olympiad Math Olympiad Videos

Circles and points | HANOI 2018

Try this beautiful problem from HANOI, 2018 based on Circles and Points.

Circles – HANOI 2018


The center of a circle and nine randomly selected points on this circle are colored in red. Every pair of those points is connected by a line segment, and every point of intersection of two line segments inside the circle is colored in red. Find the largest possible number of red points.

  • is 224
  • is 220
  • is 228
  • cannot be determined from the given information

Key Concepts


Geometry

Circles

Combination

Check the Answer


Answer: is 220.

HANOI, 2018

Geometry Revisited by Coxeter

Try with Hints


First hint

Remark that a convex quadrilateral has exactly one intersection which is the intersection of its two diagonals. Consider 9 points on the circle, which give at most \(\frac{9!}{4!5!}\)=126 intersections.

Second Hint

Considering the center and three points on the circle, there are at most \(\frac{9!}{3!6!}\)=84 intersections.

Final Step

So there are at most 126+84+10=220 red points.

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AMC-8 USA Math Olympiad

Perimeter of a circle : AMC 8 2013 Problem 25

What is the area and perimeter of a circle?


A circle is a curve which maintains same distance from a fixed point called center.

The perimeter of a circle is the length of the curve and area of a circle is portion of a plane bounded by the curve.

Try the problem


A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?

$\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi$

AMC 8 2013 Problem 25

Geometry : Perimeter of a circle

7 out of 10

Mathematical Circles.

Knowledge Graph


Perimeter of a circle- knowledge graph

Use some hints


First I want to give you the formula required.

You can clearly notice that we have to find the perimeters of all of the semicircles

The perimeter of a circle of radius $r$ unit can be obtained by the formula $2\pi r$. Then can you find perimeter of the semicircles ?!!!

So using the formula, the perimeters of

Semicircle 1 =$\frac{2\pi\times 100}{2}$ inches.

Semicircle 2 =$\frac{2\pi\times 60}{2}$ inches.

Semicircle 3 =$\frac{2\pi\times 80}{2}$ inches.

So the total path covered by the ball is

$\pi(100+60+80)=240\pi$ inches.

Is it the final answer??? Or have we ignored something ?

OK !!! please notice that they have asked for the distance covered by the center of the ball.

And the ball is of radius \(2\) inches.

So for the \(1^{st}\) and \(3^{rd}\) semicircle : The center will roll along a semicircular path of radius \(R_1-2\) and \(R_3-2\).

See this image :

And for the \(2^{nd}\) semicircle : The center will roll along a semicircular path of radius \(R_2+2\).

See the image below :

So the length of the path covered by the center of the ball is

\([\pi(100-2)+\pi(60+2)+\pi(80-2)] \quad \text{inches} \\=\pi(98+62+78) \quad \text{inches}\\=238\pi \quad \text{inches}\).

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AMC-8 USA Math Olympiad

2D Geometry – Areas related to circle AMC 8 2017 Problem 25

What are we learning ?

Competency in Focus: 2D Geometry (Areas related to circle)

This problem from American Mathematics Contest 8 (AMC 8, 2017) is based on calculation of areas related to circle. It is Question no. 25 of the AMC 8 2017 Problem series.

First look at the knowledge graph:-

calculation of  mean and median- AMC 8 2013 Problem

Next understand the problem

In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$. Arcs $TR$ and $SR$ are each one-sixth of a circle with radius 2. What is the area of the region shown? $\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$
Source of the problem
American Mathematical Contest 2017, AMC 8 Problem 25
Key Competency

Finding the area of a triangle and sector of a circle. (Area related to circles)

Difficulty Level
5/10
Suggested Book

Start with hints 

Do you really need a hint? Try it first!
C0nstruction : Let $X$ and $Y$ are the centres of the scetors $ST$ and $TR$ Now Let us join $SX$ and $TY$ What do you think? Will the points $U,S,\textbf{ and}\quad X$ be in a straightline?
$U,S,\textbf{ and}\quad X$ will be in a straight line because $\angle STU =60^{\circ}$ And angle of a  circle is $360$  i.e., $\angle SXR = \angle TYR = 60^{\circ}$ [Since sector($SXR$)=$\frac{1}{6}circle$] Then $UXY$ will make an equilateral triangle.
So after construction the figure will look like this : Therefore, The required area = Area of $\triangle UXY$ – $2 \times$ Area of the sector $SXR$.  
Area of equilateral triangle $\triangle UXY= 4\sqrt{3}$ And the are of sector $SXR= \frac{2\pi}{3}$ ANS : $4\sqrt{3}-\frac{4\pi}{3}$
Area of an equilateral triangle =$\frac{a^2\sqrt{3}}{4}$ [where $a$ is a sied of the triangle] Area of a sector of a circle of angle $\theta$ = $\frac{\theta}{360}\pi r^2$ [where $r$ is the radius of the circle]

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