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AMC 10 Geometry Math Olympiad USA Math Olympiad

Rectangular Piece of Paper | AMC 10A, 2014| Problem No 22

Try this beautiful Problem on Geometry based on Rectangular Piece of Paper from AMC 10 A, 2014. You may use sequential hints to solve the problem.

Rectangular Piece of Paper  РAMC-10A, 2014- Problem 23


A rectangular piece of paper whose length is $\sqrt{3}$ times the width has area $A$. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area $B$. What is the ratio $B: A ?$

,

  • $2:3$
  • $1:2$
  • $1:3$
  • $3:2$
  • $2:5$

Key Concepts


Geometry

Rectangle

Ratio

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Pre College Mathematics

Source of the problem

AMC-10A, 2014 Problem-23

Check the answer here, but try the problem first

$2:3$

Try with Hints


First Hint

Rectangular Piece of Paper

we have to find out the  The ratio of the area of the folded paper to that of the original paper.

At first we have to find out the midpoint of the dotted line. Now draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. Now it will form a triangle . we have to find out the area of the triangle ….

Now can you finish the problem?

Second Hint

construction of Rectangular Piece of Paper

Let us assume the width of the paper is $1$  and the length is $\sqrt 3$.Now the side length of the triangle be $\frac{2 \sqrt{3}}{3}$, $\sqrt{\left(\frac{\sqrt{3}}{3}\right)^{2}+1}=\frac{2 \sqrt{3}}{3}$, and $\sqrt{\left(\frac{\sqrt{3}}{3}\right)^{2}+1}=\frac{2 \sqrt{3}}{3}$

Now for the area of the of the Paper,

It is an equilateral triangle with height $\frac{\sqrt{3}}{3} \cdot \sqrt{3}=1$ and the area will be $\frac{\frac{2 \sqrt{3}}{3} \cdot 1}{2}=\frac{\sqrt{3}}{3}$

Therefore the area of the paper will be $ 1 . \sqrt 3=\sqrt 3$

Now Can you finish the Problem?

Third Hint

construction of Rectangular Piece of Paper

Now the area of the folded paper is $\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2 \sqrt{3}}{3}$

Therefore the ratio of the area of the folded paper to that of the original paper is $2:3$

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AMC 10 Combinatorics Math Olympiad Number Theory Probability USA Math Olympiad

Probability in Marbles | AMC 10A, 2010| Problem No 23

Try this beautiful Problem on Probability in Marbles based on smallest value AMC 10 A, 2010. You may use sequential hints to solve the problem.

Probability in Marbles – AMC-10A, 2010- Problem 23


Each of 2010 boxes in a line contains a single red marble, and for $1 \leq k \leq 2010$, the box in the $k$ th position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n)<\frac{1}{2010}$ ?

,

  • $20$
  • $22$
  • $44$
  • $45$
  • $46$

Key Concepts


Probability

Combination

Marbles

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2010 Problem-23

Check the answer here, but try the problem first

$45$

Try with Hints


First Hint

Given that Each of 2010 boxes in a line contains a single red marble, and for $1 \leq k \leq 2010$, the box in the $k$ th position also contains $k$ white marbles..

Therefore The probability of drawing a white marble from box $k$ is $\frac{k}{k+1}$ and the probability of drawing a red marble from box $k$ is $\frac{1}{k+1}$

Now can you finish the problem?

Second Hint

Also given that She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles.

Therefore we can say $P(n)=\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n-1}{n}\right) \cdot \frac{1}{n+1}=\frac{1}{n(n+1)}$

Now Can you finish the Problem?

Third Hint

Therefore the probability $\frac{1}{n(n+1)}<\frac{1}{2010}$ or $n(n+1)>2010$

Now $ n^2+n-2010>0$

Now to find out the factorization we see that $45 \times 46=2070$ and $ 44 \times 45 =1980$

As $n$ is smallest so $n=45$

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AMC 10 Math Olympiad Number Theory USA Math Olympiad

Points on a circle | AMC 10A, 2010| Problem No 22

Try this beautiful Problem on Number theory based on Points on a circle from AMC 10 A, 2010. You may use sequential hints to solve the problem.

Points on a circle – AMC-10A, 2010- Problem 22


Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?

,

  • $20$
  • $22$
  • $12$
  • $25$
  • $28$

Key Concepts


Number theory

Combination

Circle

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2010 Problem-22

Check the answer here, but try the problem first

$28$

Try with Hints


First Hint

To create a chord we have to nedd two points. Threfore three chords to create a triangle and not intersect at a single point , we have to choose six points.

Now can you finish the problem?

Second Hint

Now the condition is No three chords intersect in a single point inside the circle.Now we know that for any six points we pick, there is only 1 way to connect the points such that a triangle is formed in the circle’s interior

Now Can you finish the Problem?

Third Hint

Therefore the required answer is $ 8 \choose 6$=$28$

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Circle and Equilateral Triangle | AMC 10A, 2017| Problem No 22

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.

Circle and Equilateral Triangle  РAMC-10A, 2017- Problem 22


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? $?$

,

  • $\frac{4 \sqrt{3} \pi}{27}-\frac{1}{3}$
  • $\frac{\sqrt{3}}{2}-\frac{\pi}{8}$
  • $12$
  • $\sqrt{3}-\frac{2 \sqrt{3} \pi}{9}$
  • $\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$

Key Concepts


Geometry

Triangle

Circle

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2017 Problem-22

Check the answer here, but try the problem first

$\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$

Try with Hints


First Hint

Circle and equilateral triangle figure

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

shaded circle and equilateral triangle

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?

Second Hint

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?

Third Hint

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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Algebra AMC 10 Math Olympiad USA Math Olympiad

Linear Equation Problem | AMC 10A, 2015 | Problem No.16

Try this beautiful Problem on Algebra from the Linear equation from AMC 10 A, 2015.

Linear Equation Problem – AMC-10A, 2015- Problem 16


If $y+4=(x-2)^{2}, x+4=(y-2)^{2}$, and $x \neq y$, what is the value of $x^{2}+y^{2} ?$

,

  • $11$
  • $12$
  • $15$
  • $14$
  • $6$

Key Concepts


Algebra

Equation

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2015 Problem-16

Check the answer here, but try the problem first

$15$

Try with Hints


First Hint

Given that $y+4=(x-2)^{2}, x+4=(y-2)^{2}$ . we have to find out $x^{2}+y^{2} ?$. Now add two equations $x^{2}+y^{2}-4 x-4 y+8=x+y+8$

\(\Rightarrow x^{2}+y^{2}=5(x+y)\)

Can you find out the value \(x+y\)?

Second Hint

We can also subtract the two equations to yield the equation
$x^{2}-y^{2}-4 x+4 y=y-x$

\(\Rightarrow x^{2}-y^{2}=(x+y)(x-y)=3 x-3 y=3(x-y)\)

Therefore \((x+y)(x-y)=3 x-3 y=3(x-y)\)

\(\frac{(x+y)(x-y)}{(x-y)}=\frac{3(x-y)}{(x-y)}\) [ as\( x \neq y\)]

\(\Rightarrow (x+y)=3\)

Third Hint

Therefore \(x^2+y^2=5(x+y)=5 \times 3=15\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Side of a Quadrilateral | AMC 10A, 2009 | Problem No 12

Try this beautiful Problem on Geometry based on the side of a quadrilateral from AMC 10 A, 2009. You may use sequential hints to solve the problem.

Side of a Quadrilateral  РAMC-10A, 2009- Problem 12


In quadrilateral $A B C D, A B=5, B C=17, C D=5, D A=9,$ and $B D$ is an integer. What is $B D ?$

Quadrilateral

,

  • $11$
  • $12$
  • $13$
  • $14$
  • $15$

Key Concepts


Geometry

quadrilateral

Triangle Inequility

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2009 Problem-12

Check the answer here, but try the problem first

$13$

Try with Hints


First Hint

finding the side of a quadrilateral

We have to find out \(BD\) . Given that \(A B=5, B C=17, C D=5, D A=9,\). Notice that \(BD\) divides quadrilateral $A B C D $ into \(\triangle ABD\) and \(\triangle BCD\). Can you find out using Triangle inequality……?

Second Hint

side of a quadrilateral

We know that sum of the two sides of a triangle is greater than the Third side and difference between two sides is lesser than the third side

By the triangle inequality we have \(BD \text { < }DA+AB=9+5=14\) and \(BD \text { > }BC-CD=17-5=12\)

Third Hint

finding the length of BD

Therefore We got that $12< B D< 14$, and as we know that $B D$ is an integer, we must have $B D$=\(13\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Trapezium | AMC 10A ,2009 | Problem No 23

Try this beautiful Problem on Geometry from Trapezium  from (AMC 10 A, 2009).

Trapezium – AMC-10A, 2009- Problem 23


Convex quadrilateral $A B C D$ has $A B=9$ and $C D=12$. Diagonals $A C$ and $B D$ intersect at $E, A C=14$, and $\triangle A E D$ and $\triangle B E C$ have equal areas. What is $A E ?$

Area of trapezium

,

  • $11$
  • $12$
  • $13$
  • $14$
  • $6$

Key Concepts


Geometry

quadrilateral

Similarity

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2009 Problem-23

Check the answer here, but try the problem first

$6$

Try with Hints


First Hint

Trapezium

Given that Convex quadrilateral $A B C D$ has $A B=9$ and $C D=12$. Diagonals $A C$ and $B D$ intersect at $E, A C=14$, and $\triangle A E D$ and $\triangle B E C$ have equal areas. we have to find out the length of \(AE\).

Now if we can show that \(\triangle AEB\) and \(\triangle DEC\) are similar then we can find out \(AE\)?

Can you find out?

Second Hint

Given that area of \(\triangle AED\) and area of \(\triangle BEC\) are equal. Now area of \(\triangle ABD\) = area of \(\triangle AED\) + \(\triangle ABE\)

Area of \(\triangle ABC\) = area of \(\triangle AEB\) + \(\triangle BEC\)

Therefore area of \(\triangle ABD\)= area of \(\triangle ABC\) [as area of \(\triangle AED\) and area of \(\triangle BEC\) are equal]

Since triangles $A B D$ and $A B C$ share a base, they also have the same height and thus $\overline{A B} || \overline{C D}$ and $\triangle A E B \sim \triangle C E D$ with a ratio of 3: 4

Can you finish the problem?

Third Hint

Trapezium

Therefore $A E=\frac{3}{7} \times A C,$ so $A E=\frac{3}{7} \times 14=6$

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Circular arc | AMC 10A ,2012 | Problem No 18

Try this beautiful Problem on Geometry from Circular arc from (AMC 10 A, 2012).

Circular arc – AMC-10A, 2012- Problem 18


The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2 \pi}{3},$ where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side $2 .$ What is the area enclosed by the curve?

Circular Arc

,

  • $2 \pi+6$
  • $2 \pi+4 \sqrt{3}$
  • $3 \pi+4$
  • $2 \pi+3 \sqrt{3}+2$
  • $\pi+6 \sqrt{3}$

Key Concepts


Geometry

Circle

Hexagon

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2012

Check the answer here, but try the problem first

$\pi+6 \sqrt{3}$

Try with Hints


First Hint

Circular arc

We have to find out the area enclosed by the curve. but we can not find out eassily. Now we join the centre of 9 congruent circles. it will form a hexagon.

Circular arc with hexagon

Can you find out the required area of the closed curve? Can you finish the problem?

Second Hint

Circular arc with hexagon

The area of the Hexagon = $(6)(0.5)(2 \sqrt{3})=6 \sqrt{3}$. But we have to find out the area area enclosed by the curve. Then add the areas of the three sectors outside the hexaqon and subtract the areas of the three sectors inside the hexagon but outside the fiqure to get the area enclosed in the curved figure

Third Hint

 hexagon

The areas of the three sectors outside the hexaqon=$2 \pi$ .

the areas of the three sectors inside the hexagon but outside the figure $(\pi)$

Therefore the area enclosed by the curve is $\pi+6 \sqrt{3}$

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Area of rectangle | AMC 10A ,2012 | Problem No 21

Try this beautiful Problem on Geometry from Area of rectangle from (AMC 10 A, 2012).

Area of rectangle – AMC-10A, 2012- Problem 21


Let points $A=(0,0,0), B=(1,0,0), C=(0,2,0),$ and $D=(0,0,3)$. Points $E, F, G,$ and $H$ are midpoints of line segments $\overline{B D}, \overline{A B}, \overline{A C},$ and $\overline{D C}$ respectively. What is the area of rectangle $E F G H ?$

,

  • $\sqrt{2}$
  • $\frac{2 \sqrt{5}}{3}$
  • $\frac{3 \sqrt{5}}{4}$
  • $\sqrt{3}$
  • $\frac{2 \sqrt{7}}{3}$

Key Concepts


 Tetrahedron 

Area of rectangle

Co -ordinate geometry

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2012, Problem 21

Check the answer here, but try the problem first

$\frac{3 \sqrt{5}}{4}$

Try with Hints


First Hint

Area of rectangle

We have to find out the area of the rectangle \(EFGH\). so we have to compute the co-ordinate of the points \(E\), \(F\), \(G\), \(H\) . Next we have to find out the length of the sides \(EF\), \(FG\) , \(GH\), \(EH\). Next since rectangle area will be \(EF\) \(\times FG\)

Can you solve the problem?

Second Hint

Area of rectangle

Now co-ordinates of the points are $E(0.5,0,1.5), F(0.5,0,0), G(0,1,0), H(0,1,1.5)$. The vector $E F$ is (0,0,-1.5) , while the vector $H G$ is also (0,0,-1.5) , meaning the two sides $E F$ and $G H$ are parallel. Similarly, the vector $F G$ is (-0.5,1,0) , while the vector $E H$ is also (-0.5,1,0) . Again, these are equal in both magnitude and direction, so $F G$ and $E H$ are parallel. Thus, figure $E F G H$ is a parallelogram.The area of the Hexagon = $(6)(0.5)(2 \sqrt{3})=6 \sqrt{3}$. But we have to find out the area area enclosed by the curve. Then add the areas of the three sectors outside the hexaqon and subtract the areas of the three sectors inside the hexagon but outside the fiqure to get the area enclosed in the curved figure

Can you solve the problem?

Third Hint

Area of rectangle problem

Taking the dot product of vector $E F$ and vector $F G$ gives $0 \cdot-0.5+0 \cdot 1+-1.5 \cdot 0=0,$ which means the two vectors are perpendicular. (Alternately, as above, note that vector $E F$ goes directly down on the z-axis, while vector $F G$ has no z-component and lie completely in the xy plane.) Thus, the figure is a parallelogram with a right angle, which makes it a rectangle.

Using the distance formula we get \(E F=\frac{3}{2} \text { and } F G=\frac{\sqrt{5}}{2}\)

Therefore area of the rectangle \(EFGH\)=\(EF \times GH\)=$\frac{3}{2} \cdot \frac{\sqrt{5}}{2}$=$\frac{3 \sqrt{5}}{4}$

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