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AMC 10 Geometry Math Olympiad USA Math Olympiad

GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry based on GCF & Rectangle from AMC 10 A, 2010. You may use sequential hints to solve the problem.

GCF & Rectangle – AMC-10A, 2016- Problem 19


In rectangle $A B C D, A B=6$ and $B C=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $B E=E F=F C$. Segments $\overline{A E}$ and $\overline{A F}$ intersect $\overline{B D}$ at $P$ and $Q$, respectively. The ratio $B P: P Q: Q D$ can be written as $r: s: t$ where the greatest common factor of $r, s,$ and $t$ is $1 .$ What is $r+s+t ?$

,

  • $7$
  • $9$
  • $12$
  • $15$
  • $20$

Key Concepts


Geometry

Rectangle

Diagonal

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2016 Problem-19

Check the answer here, but try the problem first

$20$

Try with Hints


First Hint

Given that , rectangle $A B C D, A B=6$ and $B C=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that Segments $\overline{A E}$ and $\overline{A F}$ intersect $\overline{B D}$ at $P$ and $Q,$ respectively. The ratio $B P: P Q: Q D$ can be written as $r: s: t$. we have to find out $r+s+t ?$, where greatest common factor of \(r,s,t\) is \(1\)

Now $\triangle A P D \sim \triangle E P B$. From this relation we can find out a relation between \(DP\) and \(PB\)

Now can you finish the problem?

Second Hint

Now $\triangle A P D \sim \triangle E P B$\(\Rightarrow\) $\frac{D P}{P B}=\frac{A D}{B E}=3$ Therefore $P B=\frac{B D}{4}$.

SimIarly from the \(\triangle AQD \sim \triangle BQF\) \(\Rightarrow \)$\frac{D Q}{Q B}=\frac{3}{2}$

Therefore we can say that $D Q=\frac{3 \cdot B D}{5}$

Now can you finish the problem?

Third Hint

Therefore $r: s: t=\frac{1}{4}: \frac{2}{5}-\frac{1}{4}: \frac{3}{5}=5: 3: 12,$ so $r+s+t$=\(20\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Geometry Problem based on a fly trapped inside cubical box from AMC 10 A, 2010. You may use sequential hints to solve the problem.

Fly trapped inside cubical box – AMC-10A, 2010- Problem 20


A fly trapped inside a cubical box with side length 1 meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?

,

  • $4+4 \sqrt{2}$
  • $2+4 \sqrt{2}+2 \sqrt{3}$
  • $2+3 \sqrt{2}+3 \sqrt{3}$
  • $4 \sqrt{2}+4 \sqrt{3}$
  • $3 \sqrt{2}+5 \sqrt{3}$

Key Concepts


Geometry

Cube

Diagonal

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Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2010 Problem-20

Check the answer here, but try the problem first

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

Try with Hints


First Hint

Suppose the fly starts from the point \(A\).we have to find out  the maximum possible length. The maximum possible length will be from one corner to another corner such as …..

$A \rightarrow G \rightarrow B \rightarrow H \rightarrow C \rightarrow E \rightarrow D \rightarrow F \rightarrow A$

Now can you find out this maximum path?

Now can you finish the problem?

Second Hint

Given that the side length of the cube is $1$. Therefore the diagonal \(AC\)=\(\sqrt 2\) and the diagonal \(AG=\sqrt 3\). Now we have to find out the path \(AG+GB+BH+HC+CE+ED+DF+FA\).Can you find it ?

Now can you finish the problem?

Third Hint

\(AG+GB+BH+HC+CE+ED+DF+FA\)=\(\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2\)=(\(4\sqrt3+4\sqrt2\))

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Measure of angle | AMC 10A, 2019| Problem No 13

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

Measure of angle  – AMC-10A, 2019- Problem 13


Let $\triangle A B C$ be an isosceles triangle with $B C=A C$ and $\angle A C B=40^{\circ} .$ Construct the circle with diameter $\overline{B C}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{A C}$ and $\overline{A B}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $B C D E .$ What is the degree measure of $\angle B F C ?$

,

  • $90$
  • $100$
  • $105$
  • $110$
  • $120$

Key Concepts


Geometry

Circle

Triangle

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2019 Problem-`13

Check the answer here, but try the problem first

$110^{\circ}$

Try with Hints


First Hint

According to the questation we draw the diagram. we have to find out \(\angle BFC\)

Now \(\angle BEC\) = \(\angle BDC\) =\(90^{\circ}\) (as they are inscribed in a semicircle)

$\angle A C B=40^{\circ} .$ Therefore we can say that \(\angle ABC=70^{\circ}\) (as $\triangle A B C$ be an isosceles triangle with $B C=A C$)

Can you find out the value of $\angle B F C ?$

Now can you finish the problem?

Second Hint

As \(\angle ABC=70^{\circ}\) and \(\angle BEC=90^{\circ}\) Therefore $\angle E C B=20^{\circ}$( as sum of the angles of a triangle is\( 180^{\circ}\)

Similarly $\angle D B C=50^{\circ}$

Now Can you finish the Problem?

Third Hint

Now $\angle B D C+\angle D C B+\angle D B C=180^{\circ} \Longrightarrow 90^{\circ}+40^{\circ}+\angle D B C=180^{\circ} \Longrightarrow \angle D B C$=$50^{\circ}$

$\angle B E C+\angle E B C+\angle E C B=180^{\circ} \Longrightarrow 90^{\circ}+70^{\circ}+\angle E C B=180^{\circ} \Rightarrow \angle E C B$=$20^{\circ}$

we take triangle $B F C$, and find $\angle B F C=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$

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AMC 10 Combinatorics Math Olympiad USA Math Olympiad

Recursion Problem | AMC 10A, 2019| Problem No 15

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

Recursion- AMC-10A, 2019- Problem 15


A sequence of numbers is defined recursively by $a_{1}=1, a_{2}=\frac{3}{7},$ and $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

  • $2020$
  • $4039$
  • $6057$
  • $6061$
  • $8078$

Key Concepts


Algebra

Recursive formula

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2019 Problem-15

Check the answer here, but try the problem first

$8078$

Try with Hints


First Hint

The given expression is $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$ and given that $a_{1}=1, a_{2}=\frac{3}{7}$. we have to find out \(a_{2019}\)?

at first we may use recursive formula we can find out \(a_3\) , \(a_4\) with the help of \(a_1\), \(a_2\). later we can find out \(a_n\)

Now can you finish the problem?

Second Hint

Given that $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

Now \(n=3\) then $a_{3}=\frac{a_{(3-2)} \cdot a_{(3-1)}}{2 a_{(3-2)}-a_{(3-1)}}$

\(\Rightarrow\) $a_{3}=\frac{a_{(1)} \cdot a_{(2)}}{2 a_{(1)}-a_{(2)}}$

\(\Rightarrow\) $a_{3}=\frac{a_{(1)} \cdot a_{(2)}}{2 a_{(1)}-a_{(2)}}$

\(\Rightarrow\) $a_{3}=\frac{1*\frac{3}{7}}{2*1-\frac{3}{7}}$

\(\Rightarrow\) $a_{3}=\frac{3}{7}$

Similarly if we put \(n=4\) we get \(a_4=\frac{3}{15}\) (where $a_{1}=1, a_{2}=\frac{3}{7}$,\(a_3=\frac{3}{7}\))

Continue this way we $a_{n}=\frac{3}{4 n-1}$

So can you find out the value of \(a_{2019}\)?

Now Can you finish the Problem?

Third Hint

Now $a_{n}=\frac{3}{4 n-1}$

Put \(n=2019\)

$a_{2019}=\frac{3}{8075}$ which is the form of \(\frac{p}{q}\)

Therefore \(p+q=8078\)

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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

Suggested Book | Source | Answer


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

Suggested Book | Source | Answer


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

Suggested Book | Source | Answer


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

Suggested Book | Source | Answer


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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