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AMC 10 Combinatorics Math Olympiad USA Math Olympiad

Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra based on Roots of Polynomial from AMC 10 A, 2019. You may use sequential hints to solve the problem.

Algebra- AMC-10A, 2019- Problem 24


Let $p, q,$ and $r$ be the distinct roots of the polynomial $x^{3}-22 x^{2}+80 x-67$. It is given that there exist real numbers $A, B$, and $C$ such that $\frac{1}{s^{3}-22 s^{2}+80 s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r}$

for all $s \notin{p, q, r} .$ What is $\frac{1}{A}+\frac{1}{B}+\frac{1}{C} ?$

,

  • $243$
  • $244$
  • $245$
  • $246$
  • $247$

Key Concepts


Algebra

Linear Equation

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2019 Problem-24

Check the answer here, but try the problem first

$244$

Try with Hints


First Hint

The given equation is $\frac{1}{s^{3}-22 s^{2}+80 s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r}$…………………(1)

If we multiply both sides we will get

Multiplying both sides by $(s-p)(s-q)(s-r)$ we will get
$$
1=A(s-q)(s-r)+B(s-p)(s-r)+C(s-p)(s-q)
$$

Now can you finish the problem?

Second Hint

Now Put $S=P$ we will get $\frac{1}{A}=(p-q)(p-r)$…………(2)

Now Put $S=q$ we will get $\frac{1}{B}=(q-p)(q-r)$………..(3)

Now Put $S=r$ we will get $\frac{1}{C}=(r-p)(r-q)$………..(4)

Now Can you finish the Problem?

Third Hint

Adding (2) +(3)+(4) we get,$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=p^{2}+q^{2}+r^{2}-p q-q r-p r$

Now Using Vieta’s Formulas, $p^{2}+q^{2}+r^{2}=(p+q+r)^{2}-2(p q+q r+p r)=324$ and $p q+q r+p r=80$

Therefore the required answer is $324-80$=$244$

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra based on Set of Fractions from AMC 10 A, 2015. You may use sequential hints to solve the problem.

Set of Fractions – AMC-10A, 2015- Problem 15


Consider the set of all fractions $\frac{x}{y}$, where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1,$ the value of the fraction is increased by $10 \% ?$

,

  • $0$
  • $1$
  • $2$
  • $3$
  • $infinitely many$

Key Concepts


Algebra

fraction

factorization

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2015 Problem-15

Check the answer here, but try the problem first

$1$

Try with Hints


First Hint

According to the questation we can write $\frac{x+1}{y+1}=\frac{11 x}{10 y}$

\(\Rightarrow xy +11x-10y=0\)

\(\Rightarrow (x-10)(y-11)=-110\)

Now can you finish the problem?

Second Hint

Here \(x\) and \(y\) must positive, so $x>0$ and $y>0$, so $x-10>-10$ and $y+11>11$

Now we have to find out the factors of \(110\) and find out the possible pairs to fulfill the condition….

Now Can you finish the Problem?

Third Hint

uses the factors of $110$ , we can get the factor pairs: $(-1,110),(-2,55),$ and $(-5,22)$
But we can’t stop here because $x$ and $y$ must be relatively prime.
$(-1,110 )$ gives $x=9$ and $y=99.9$ and 99 are not relatively prime, so this doesn’t work.
$(-2,55 )$ gives $x=8$ and $y=44$. This doesn’t work.
$(-5,22)$ gives $x=5$ and $y=11$. This does work.

Therefore the one solution exist

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Geometry based on Positive Integers and Quadrilateral from AMC 10 A, 2015. You may use sequential hints to solve the problem.

Positive Integers and Quadrilateral – AMC-10A, 2015- Problem 24


For some positive integers $p$, there is a quadrilateral $A B C D$ with positive integer side lengths, perimeter $p$, right angles at $B$ and $C, A B=2$, and $C D=A D$. How many different values of $p<2015$ are possible?

,

  • $30$
  • $31$
  • $61$
  • $62$
  • $63$

Key Concepts


Geometry

Rectangle

Integer

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2015 Problem-24

Check the answer here, but try the problem first

$31$

Try with Hints


First Hint

Positive Integers and Quadrilateral

Given that $ ABCD$ is a quadrilateral whose perimeter $p$, right angles at $B$ and $C, A B=2$, and $C D=A D$. we have to find out How many different values of $p<2015$ are possible.

Now draw a perpendicular $AE$ on $CD$ . Let us assume that $BC=AE=x$ Then $CE=2$ and $DE=y-2$

Now can you finish the problem?

Second Hint

Positive Integers and Quadrilateral

Now from the \(\triangle ADE\) we can write $x^{2}+(y-2)^{2}=y^{2}$

\(\Rightarrow x^{2}-4 y+4=0\)

\(\Rightarrow x^2=4(y-1)\), Thus, $y$ is one more than a perfect square.

Therefore the perimeter will be $p=2+x+2 y=2 y+2 \sqrt{y-1}+2$

Now according to the problem $p<2015$

So, $p=2+x+2 y=2 y+2 \sqrt{y-1}+2 <2015$

Now Can you finish the Problem?

Third Hint


Quadrilateral

Now $y=31^{2}+1=962$ and $y=32^{2}+1=1025$

Here $y=31^{2}+1=962$ is valid but $y=32^{2}+1=1025$ is not. On the lower side, $y=1$ does not work (because $x>0$ ), but $y=1^{2}+1$ does work. Hence, there are $31$ valid $y$ (all $y$ such that $y=n^{2}+1$ for $1 \leq n \leq 31$ )

Therefore the correct answer is $31$

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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  – AMC-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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Suggested Reading

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  – AMC-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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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  – AMC-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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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

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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