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AMC-8 Math Olympiad USA Math Olympiad

Median of numbers | AMC-10A, 2020 | Problem 11

Try this beautiful problem from Geometry based on Median of numbers from AMC 10A, 2020.

Median of numbers – AMC-10A, 2020- Problem 11


What is the median of the following list of $4040$ numbers$?$

\(1,2,3,…….2020,1^2,2^2,3^2………..{2020}^2\)

  • $1989.5$
  • $ 1976.5$
  • $1972.5$

Key Concepts


Median

Algebra

square numbers

Check the Answer


Answer: $1976.5$

AMC-10A (2020) Problem 11

Pre College Mathematics

Try with Hints


To find the median we need to know how many terms are there and the position of the numbers .here two types of numbers, first nonsquare i.e (1,2,3……2020) and squares numbers i.e \((1^2,2^2,3^2……2020^2)\).so We want to know the \(2020\)th term and the \(2021\)st term to get the median.

Can you now finish the problem ……….

Now less than 2020 the square number is \({44}^2\)=1936 and if we take \({45}^2\)=2025 which is greater than 2020.therefore we take the term that \(1,2,3…2020\) trms + 44 terms=\(2064\) terms.

can you finish the problem……..

since $44^{2}$ is $44+45=89$ less than $45^{2}=2025$ and 84 less than 2020 we will only need to consider the perfect square terms going down from the 2064 th term, 2020, after going down $84$ terms. Since the $2020$th and $2021$st terms are only $44$ and $43$ terms away from the $2064$th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$. Averaging the two,=\(1976.5\)

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AMC 10 Math Olympiad USA Math Olympiad

Triangle Area Problem | AMC-10A, 2009 | Problem 10

Try this beautiful problem from Geometry based on Area of Triangle.

Area of Triangle – AMC-10A, 2009- Problem 10


Triangle $A B C$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B, A D=3$, and $D C=4 .$ What is the area of $\triangle A B C$ ?

area of triangle - problem figure
Area of Triangle Problem
  • $4 \sqrt{3}$
  • $7 \sqrt{3}$
  • $14 \sqrt{3}$
  • \(21\)
  • \(42\)

Key Concepts


Triangle

Similarity

Geometry

Check the Answer


Answer: $7 \sqrt{3}$

AMC-10A (2009) Problem 10

Pre College Mathematics

Try with Hints


area of triangle - problem

We have to find out the area of the Triangle ABC where \(\angle B=90^{\circ}\) and \(BD \perp AC\)

Area of a Triangle = \(\frac{1}{2}\times \) Base \(\times\) Height.But we don know the value of \(AB\) & \(BC\). But we know \(AC=7\). So if we can find out the value of \(BD\) then we can find out the are of \(\triangle ABC\) by \(\frac{1}{2}\times AC \times BD\)

Can you now finish the problem ……….

area of triangle - problem

Let \(\angle C=\theta\), then \(\angle A=(90-\theta)\) (as \(\angle B=90^{\circ}\), Sum of the angles in a triangle is \(180^{\circ}\))

In \(\triangle ABD\), \(\angle ABD=\theta\) \(\Rightarrow \angle A=(90-\theta\))

Again In \(\triangle DBC\), \(\angle DBC\)=(\(90-\theta\)) \(\Rightarrow \angle C=\theta\)

From the above condition we say that , \(\triangle ABD \sim \triangle BDC\)

Therefore , \(\frac{BD}{CD}=\frac{AD}{BD}\) \(\Rightarrow {BD}^2=AD.CD=4\times 3\)

\(\Rightarrow BD=\sqrt {12}\)

can you finish the problem……..

area of triangle

Therefore area of the \(\triangle ABC=\frac{1}{2}\times AC \times BD=\frac{1}{2}\times 7 \times \sqrt{12}=7 \sqrt{3}\)

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AMC 10 Math Olympiad USA Math Olympiad

Side Length of Rectangle | AMC-10A, 2009 | Problem 17

Try this beautiful problem from Geometry based on Side Length of Rectangle.

Side Length of Rectangle – AMC-10A, 2009- Problem 17


Rectangle $A B C D$ has $A B=4$ and $B C=3 .$ Segment $E F$ is constructed through $B$ so that $E F$ isperpendicular to $D B$, and $A$ and $C$ lie on $D E$ and $D F$, respectively. What is $E F$ ?

  • $9$
  • $10$
  • $\frac{125}{12}$
  • \(\frac{103}{9}\)
  • \(12\)

Key Concepts


Triangle

Rectangle

Geometry

Check the Answer


Answer: $\frac{125}{12}$

AMC-10A (2009) Problem 10

Pre College Mathematics

Try with Hints


Finding Side Length of Rectangle

We have to find out the length of \(EF\)

Now $BD$ is the altitude from $B$ to $EF$, we can use the equation $BD^2 = EB\cdot BF$. ( as \(\triangle BDE \sim \triangle BDF\)).so we have to find out \(BE\) and \(BF\)

Can you now finish the problem ……….

Problem figure

Now Clearly, $\triangle BDE \sim \triangle DCB$. Because of this, $\frac{A B}{C B}=\frac{E B}{D B}$. From the given information and the Pythagorean theorem, $A B=4, C B=3$, and $D B=5 .$ Solving gives $E B=20 / 3$
We can use the above formula to solve for $B F . B D^{2}=20 / 3 \cdot B F$. Solve to obtain $B F=15 / 4$

can you finish the problem……..

Problem figure

Therefore $E F=E B+B F=\frac{20}{3}+\frac{15}{4}=\frac{80+45}{12}$

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AMC-8 Geometry Math Olympiad USA Math Olympiad

The area of trapezoid | AMC 8, 2003 | Problem 21

Try this beautiful problem from Geometry: The area of trapezoid

The area of trapezoid – AMC-8, 2003- Problem 21


The area of trapezoid ABCD is 164 \(cm^2\). The altitude is  8 cm, AB is 10 cm, and CD is 17 cm. What is BC in centimeters?

The area of trapezoid - problem figure

,

 i

  • $8$
  • $ 10 $
  • $15$

Key Concepts


Geometry

trapezoid

Triangle

Check the Answer


Answer: $ 10 $

AMC-8 (2003) Problem 21

Pre College Mathematics

Try with Hints


Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Can you now finish the problem ……….

finding the area of trapezoid

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

can you finish the problem……..

finding the area of trapezoid

Given that the area of the trapezoid is 164 sq.unit

Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

Using Pythagorean rules on the triangle ABD,we have…

\((AD)^2 + (BD)^2 =(AB)^2\)

\( \Rightarrow (AD)^2 + (8)^2 =(10)^2\)

\( \Rightarrow (AD)^2 =(10)^2 – (8)^2 \)

\( \Rightarrow (AD)^2 = 36\)

\( \Rightarrow (AD) =6\)

Using Pythagorean rules on the triangle CED,we have…

\((CE)^2 + (DE)^2 =(DC)^2\)

\( \Rightarrow (CE)^2 + (8)^2 =(17)^2\)

\( \Rightarrow (CE)^2 =(17)^2 – (8)^2 \)

\( \Rightarrow (CE)^2 = 225\)

\( \Rightarrow (CE) =15\)

Let BC= DE=x

Therefore area of the trapezoid=\(\frac{1}{2} \times (AD+BC) \times 8\)=164

\(\Rightarrow \frac{1}{2} \times (6+x+15) \times 8\) =164

\(\Rightarrow x=10\)

Therefore BC=10 cm

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AMC-8 Math Olympiad USA Math Olympiad

Largest area Problem | AMC 8, 2003 | Problem 22

Try this beautiful problem from Geometry based Largest area.

Largest area – AMC-8, 2003 – Problem 22


The following figures are composed of squares and circles. Which figure has a shaded region with largest area?

Problem figure
  • $A$
  • $B$
  • $C$

Key Concepts


Geometry

Circle

Square

Check the Answer


Answer:$C$

AMC-8 (2003) Problem 22

Pre College Mathematics

Try with Hints


To find out the largest area at first we have to find out the radius of the circles . all the circles are inscribed ito the squares .now there is a relation between the radius and the side length of the squares….

Can you now finish the problem ……….

area of circle =\(\pi r^2\)

can you finish the problem……..

Largest  area Problem

In A:

Total area of the square =\(2^2=4\)

Now the radius of the inscribed be 1(as the diameter of circle = side length of the side =2)

Area of the inscribed circle is \(\pi (1)^2=\pi\)

Therefore the shaded area =\(4- \pi\)

In B:

Largest  area Problem - figure

Total area of the square =\(2^2=4\)

There are 4 circle and radius of one circle be \(\frac{1}{2}\)

Total area pf 4 circles be \(4 \times \pi \times (\frac{1}{2})^2=\pi\)

Therefore the shaded area =\(4- \pi\)

In C:

finding the largest area

Total area of the square =\(2^2=4\)

Now the length of the diameter = length of the diagonal of the square=2

Therefore radius of the circle=\(\pi\) and lengthe of the side of the square=\(\sqrt 2\)

Thertefore area of the shaded region=Area of the square-Area of the circle=\(\pi (1)^2-(\sqrt 2)^2\)=\(\pi – 2\)

Therefore the answer is  C

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AMC 10 Math Olympiad USA Math Olympiad

Greatest Common Divisor | AMC-10A, 2018 | Problem 22

Try this beautiful problem from Algebra based on Greatest Common Divisor from AMC 10A, 2018, Problem 22.

Greatest Common Divisor – AMC-10A, 2018- Problem 22


Let \(a, b, c,\) and \(d\) be positive integers such that \(\gcd(a, b)=24\), \(\gcd(b, c)=36\), \(\gcd(c, d)=54\), and \(70<\gcd(d, a)<100\). Which of the following must be a divisor of \(a\)?

  • \(5\)
  • \(7\)
  • \(13\)

Key Concepts


Number theory

Gcd

Divisior

Check the Answer


Answer: \(13\)

AMC-10A (2018) Problem 22

Pre College Mathematics

Try with Hints


TO find the divisor of \(a\) at first we have to find the value of \(a\).can you find the value of \(a\)?

Given that \(\gcd(a, b)=24\), \(\gcd(b, c)=36\), \(\gcd(c, d)=54\), and \(70<\gcd(d, a)<100\)

so we can say \(a=24 \times\) some integer and \(b=24 \times\) some another integer (according to gcd rules)

similarly for the others c & d…..

now if we can find out the value of \(\gcd(d, a)\) then we may use the condition \(70<\gcd(d, a)<100\)

Can you now finish the problem ……….

so we may say that \(gcd(a, b)\) is \(2^3 * 3\) and the \(gcd\) of \((c, d)\) is \(2 * 3^3\). However, the \(gcd\) of \((b, c) = 2^2 * 3^2\) (meaning both are divisible by 36). Therefore, \(a\) is only divisible by \(3^1\) (and no higher power of 3), while \(d\) is divisible by only \(2^1\) (and no higher power of 2).

can you finish the problem……..

so we can say that \(gcd\) of \((a, d)\) can be expressed in the form \(2 \times 3 \times \) some positve integer and now \(k\) is a number not divisible by \(2\) or \(3\). so from the given numbes it will be \(13\) because \(2 \times 3 \times k\) must lie \(70<\gcd (d, a)<100\). so the required ans is \(13\)

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AMC-8 Math Olympiad USA Math Olympiad

Area of the Trapezium | AMC-10A, 2018 | Problem 24

Try this beautiful problem from Geometry based on the Area of the Trapezium.

Area of the Trapezium – AMC-10A, 2018- Problem 24


Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

  • $79$
  • $ 75$
  • $82$

Key Concepts


Geometry

Triangle

Trapezium

Check the Answer


Answer: $75$

AMC-10A (2018) Problem 24

Pre College Mathematics

Try with Hints


Area of the Trapezium

We have to find out the area of BGFD.Given that AG is the angle bisector of \(\angle BAC\) ,\(D\) and \(E\) are the mid points of \(AB\) and \(AC\). so we may say that \(DE ||BC\) by mid point theorm…

So clearly BGFD is a Trapezium.now area of the trapezium=\(\frac{1}{2} (BG+DF) \times height betwween DF and BG\)

can you find out the value of \(BG,DF \) and height between them….?

Can you now finish the problem ……….

Area of the Trapezium- Problem

Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$.

Therefore area of \(\triangle ABC\)=\(\frac{ah}{2}\)=\(120\)………………..(1)

From the angle bisector theorem, we have that\(\frac{50}{x} = \frac{10}{y}\) i.e \(\frac{x}{y}=5\)

Let \(BC\)=\(a\) then \(BG\)=\(\frac{5a}{6}\) and \(DF\)=\(\frac{1}{2 } \times BG\) i.e \(\frac{5a}{12}\)

now can you find out the area of Trapezium and area of Triangle?

can you finish the problem……..

Area of the shaded portion

Therefore area of the Trapezium=\(\frac{1}{2} (BG+DF) \times FG\)=\(\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}\)=\(\frac{ah}{2} \times \frac{15}{24}\)=\(120 \times \frac{15}{24}\)=\(75\) \((from ……..(1))\)

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AMC-8 Math Olympiad USA Math Olympiad

Measuring the length in Triangle | AMC-10B, 2011 | Problem 9

Try this beautiful problem from Geometry and solve it by measuring the length in triangle.

Measuring the length in Triangle- AMC-10B, 2011- Problem 9


The area of $\triangle$$EBD$ is one third of the area of $\triangle$$ABC$. Segment $DE$ is perpendicular to segment $AB$. What is $BD$?

Measuring the length in Triangle- Problem
  • \(8\sqrt 3\)
  • \(\frac {4\sqrt3}{3}\)
  • \(6\sqrt 3\)

Key Concepts


Geometry

Triangle

similarity

Check the Answer


Answer: \(\frac{ 4\sqrt 3}{3}\)

AMC-10B (2011) Problem 9

Pre College Mathematics

Try with Hints


Find BD

We have to find out the length of \(BD\). The given informations are “The area of $\triangle$$EBD$ is one third of the area of $\triangle$$ABC$. Segment $DE$ is perpendicular to segment $AB$”

If you notice very carefully about the side lengths of the \(\triangle ABC\) then \(AC=3,BC=4,AB=5\) i.e \((AC)^2+(AB)^2=(3)^2+(4)^2=25=(AB)^2\)……..So from the pythagorean theorm we can say that \(\angle ACB=90^{\circ} \)

Therefore area of \(\triangle ACB=\frac{1}{2} \times 3 \times 4=6\)

so area of the \(\triangle BDE=\frac{1}{3} \times 6=2\)

Now the \(\triangle BDE\) and \(\triangle ABC\) If we can show that two triangles are similar then we will get the value of \(BD\).Can you prove \(\triangle BDE \sim \triangle ABC\) ?

Can you now finish the problem ……….

Finding the measurement

In \(\triangle BDE\) & \(\triangle ACB\) we have…..

\(\angle B=X\) \(\Rightarrow \angle BED=(90-x)\) and \(\angle CAB=(90-X)\) (AS \(\angle ACB=90\) & sum of the angles of a triangle is 180)

Therefore \(\triangle BDE \sim \triangle BCD\)

can you finish the problem……..

The value of BD:

Now \(\triangle BDE \sim \triangle BCD\) \(\Rightarrow \frac{(BD)^2}{(BC)^2}=\frac{Area of \triangle BDE}{Area of triangle ACB}\) =\(\frac{(BD)^2}{16}=\frac{2}{6}\)

So \(BD=\frac{ 4\sqrt 3}{3}\)

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AMC-8 Math Olympiad USA Math Olympiad

Problem based on LCM | AMC 8, 2016 | Problem 20

Try this beautiful problem from Algebra based on LCM from AMC-8, 2016.

Problem based on LCM – AMC 8, 2016


The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

  • \(26\)
  • \(20\)
  • \(28\)

Key Concepts


Algebra

Divisor

multiplication

Check the Answer


Answer:20

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


We have to find out the least common multiple of $a$ and $c$.if you know the value of \(a\) and \(c\) then you can easily find out the required LCM. Can you find out the value of \(a\) and \(c\)?

Can you now finish the problem ……….

Given that the least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$ .then b must divide 12 and 15. There is only one possibility that b=3 which divide 12 and 15. therefore \(a\)=\(\frac{12}{3}=4\)

can you finish the problem……..

so\(b\)=3. Given that LCM of \(b\) and \(c\) is 15. Therefore c=5

Now lcm of \(a\) and \(c\) that is lcm of 4 and 5=20

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AMC-8 Math Olympiad USA Math Olympiad

Ratio of LCM & GCF | Algebra | AMC 8, 2013 | Problem 10

Try this beautiful problem from Algebra based on the ratio of LCM & GCF from AMC-8, 2013.

Ratio of LCM & GCF | AMC-8, 2013 | Problem 10


What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?

  • 310
  • 330
  • 360

Key Concepts


Algebra

Ratio

LCM & GCF

Check the Answer


Answer:$330$

AMC-8, 2013 problem 10

Challenges and Thrills in Pre College Mathematics

Try with Hints


We have to find out the ratio of least common multiple and greatest common factor of 180 and 594. So at first, we have to find out prime factors of 180 & 594. Now…….

\(180=3^2\times 5 \times 2^2\)

\(594=3^3 \times 11 \times 2\)

Can you now finish the problem ……….

Now lcm of two numbers i.e multiplications of the greatest power of all the numbers 

Therefore LCM of 180 & 594=\(3^3\times2^2 \times 11 \times 5\)=\(5940\)

For the GCF of 180 and 594, multiplications of the least power of all of the numbers i.e \(3^2\times 2\)=\(18\)

can you finish the problem……..

Therefore the ratio of Lcm & gcf of 180 and 594 =\(\frac{5940}{18}\)=\(330\)

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