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AMC 10 Geometry Math Olympiad USA Math Olympiad

Problem on Circle and Triangle | AMC 10A, 2016 | Problem 21

Try this beautiful problem from Geometry: Problem on Circle and Triangle

Problem on Circle and Triangle – AMC-10A, 2016- Question 21


Circles with centers $P, Q$ and $R,$ having radii 1,2 and 3 , respectively, lie on the same side of line $l$ and are tangent to $l$ at $P^{\prime}, Q^{\prime}$ and $R^{\prime}$ respectively, with $Q^{\prime}$ between $P^{\prime}$ and $R^{\prime}$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $P Q R ?$

,

 

  • $0$
  • $\sqrt{6} / 3$
  • $1$
  • $\sqrt{6}-\sqrt{2}$
  • $\sqrt{6} / 2$

Key Concepts


Geometry

Circle

Triangle

Check the Answer


But try the problem first…

Answer: $\sqrt{6}-\sqrt{2}$

Source
Suggested Reading

AMC-10A (2016) Problem 21

Pre College Mathematics

Try with Hints


First hint

We have to find out area of the Triangle PQR. But PQR is not a Standard Triangle that we can find out eassily. Join $PP^{\prime}$, $QQ^{\prime}$, $RR^{\prime}$. Now we can find out PQR such that $\left[P^{\prime} P Q R R^{\prime}\right]$ in two different ways: $\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]$ and $[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$, so $\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]=[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$

Can you now finish the problem ……….

Second Hint

$P^{\prime} Q^{\prime}=\sqrt{P Q^{2}-\left(Q Q^{\prime}-P P^{\prime}\right)^{2}}=\sqrt{9-1}=\sqrt{8}=2 \sqrt{2}$

$Q^{\prime} R^{\prime}=\sqrt{Q R^{2}-\left(R R^{\prime}-Q Q^{\prime}\right)^{2}}=\sqrt{5^{2}-1^{2}}=\sqrt{24}=2 \sqrt{6}$

$\left[P^{\prime} P Q Q^{\prime}\right]=\frac{P^{\prime} P+Q^{\prime} Q}{2} * 2 \sqrt{2}=\frac{1+2}{2} * 2 \sqrt{2}=3 \sqrt{2}$

$\left[Q^{\prime} Q R R^{\prime}\right]=5 \sqrt{6}$

$\left[P^{\prime} P R R^{\prime}\right]$ = $P^{\prime} R^{\prime}=P^{\prime} Q^{\prime}+Q^{\prime} R^{\prime}=2 \sqrt{2}+2 \sqrt{6}$

$\left[P^{\prime} P R R^{\prime}\right]=4 \sqrt{2}+4 \sqrt{6}$

$\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]=[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$

$3 \sqrt{2}+5 \sqrt{6}=4 \sqrt{2}+4 \sqrt{6}+[P Q R]$

$[P Q R]=\sqrt{6}-\sqrt{2}$

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Triangle and integers | AIME I, 1995 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.

Triangle and integers – AIME I, 1995


Triangle ABC is isosceles, with AB=AC and altitude AM=11, suppose that there is a point D on AM with AD=10 and \(\angle BDC\)=3\(\angle BAC\). then the perimeter of \(\Delta ABC\) may be written in the form \(a+\sqrt{b}\) where a and b are integers, find a+b.

Triangle and integers
  • is 107
  • is 616
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Triangle

Trigonometry

Check the Answer


But try the problem first…

Answer: is 616.

Source
Suggested Reading

AIME I, 1995, Question 9

Plane Trigonometry by Loney

Try with Hints


First hint

Let x= \(\angle CAM\)

\(\Rightarrow \angle CDM =3x\)

\(\Rightarrow \frac{tan3x}{tanx}=\frac{\frac{CM}{1}}{\frac{CM}{11}}\)=11 [by trigonometry ratio property in right angled triangle]

Second Hint

\(\Rightarrow \frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx\)

solving we get, tanx=\(\frac{1}{2}\)

\(\Rightarrow CM=\frac{11}{2}\)

Final Step

\(\Rightarrow 2(AC+CM)\) where \(AC=\frac{11\sqrt {5}}{2}\) by Pythagoras formula

=\(\sqrt{605}+11\) then a+b=605+11=616.

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AIME I Geometry Math Olympiad USA Math Olympiad

Inscribed circle and perimeter | AIME I, 1999 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Inscribed circle and perimeter.

Inscribed circle and perimeter – AIME I, 1999


The inscribed circle of triangle ABC is tangent to AB at P, and its radius is 21 given that AP=23 and PB=27 find the perimeter of the triangle

Inscribed circle and perimeter
  • is 107
  • is 345
  • is 840
  • cannot be determined from the given information

Key Concepts


Inscribed circle

Perimeter

Triangle

Check the Answer


But try the problem first…

Answer: is 345.

Source
Suggested Reading

AIME I, 1999, Question 12

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

Q tangency pt on AC, R tangency pt on BC AP=AQ=23 BP=BR=27 CQ=CR=x and

Second Hint

\(s \times r =A\) and \(s=\frac{27 \times 2+23 \times 2+x \times 2}{2}=50+x\) and A=\(({(50+x)(x)(23)(27)})\) then from these equations 441(50+x)=621x then x=\(\frac{245}{2}\)

Final Step

perimeter 2s=2(50+\(\frac{245}{2}\))=345.

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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


But try the problem first…

Answer: $1976.5$

Source
Suggested Reading

AMC-10A (2020) Problem 11

Pre College Mathematics

Try with Hints


First hint

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 ……….

Second Hint

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……..

Final Step

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 Geometry Math Olympiad USA Math Olympiad

Side of Square | AMC 10A, 2013 | Problem 3

Try this beautiful problem from Geometry: Side of Square.

Sides of Square – AMC-10A, 2013- Problem 3


Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$?

,

 i

Side of Square - Problem
  • $4$
  • $5$
  • $6$
  • $7$
  • \(8\)

Key Concepts


Geometry

Square

Triangle

Check the Answer


But try the problem first…

Answer: $8$

Source
Suggested Reading

AMC-10A (2013) Problem 3

Pre College Mathematics

Try with Hints


First hint

Side of Square

Given that Square $ABCD$ has side length $10$ and area of $\triangle ABE$ is $40$.we have to find out length of \(BE\) where \(E\) is the point on \(BC\). we know area of the \(\triangle ABE=\frac{1}{2} AB.BE=40\)

Can you find out the side length of \(BE\)?

Can you now finish the problem ……….

Second Hint

Side of Square

\(\triangle ABE=\frac{1}{2} AB.BE=40\)

\(\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40\)

\(\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40\)

\(\Rightarrow BE=8\)

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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


But try the problem first…

Answer: $7 \sqrt{3}$

Source
Suggested Reading

AMC-10A (2009) Problem 10

Pre College Mathematics

Try with Hints


First hint

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 ……….

Second Hint

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……..

Final Step

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


But try the problem first…

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

Source
Suggested Reading

AMC-10A (2009) Problem 10

Pre College Mathematics

Try with Hints


First hint

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 ……….

Second Hint

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……..

Final Step

Problem figure

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

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Area of Triangle | AMC 10A, 2006 | Problem 21

Try this beautiful problem from Geometry: Area of a triangle

Triangle – AMC-10A, 2006- Problem 21


A circle of radius 1 is tangent to a circle of radius 2 . The sides of $\triangle A B C$ are tangent to the circles as shown, and the sides $\overline{A B}$ and $\overline{A C}$ are congruent. What is the area of $\triangle A B C ?$

,

 i

Area of Triangle Problem
  • $15 \sqrt{2}$
  • $\frac{35}{2} $
  • $\frac{64}{3}$
  • $16 \sqrt{2}$
  • \(24\)

Key Concepts


Geometry

Circle

Triangle

Check the Answer


But try the problem first…

Answer: $16 \sqrt{2}$

Source
Suggested Reading

AMC-10A (2006) Problem 21

Pre College Mathematics

Try with Hints


First hint

Area of Triangle - figure

Given that there are two circle of radius 1 is tangent to a circle of radius 2.we have to find out the area of the \(\triangle ABC\).Now draw a perpendicular line \(AF\) on \(BC\).Clearly it will pass through two centers \(O_1\) and \(O_2\). and $\overline{A B}$ and $\overline{A C}$ are congruent i.e \(\triangle ABC\) is an Isosceles triangle. Therefore \(BF=FC\)

So if we can find out \(AF\) and \(BC\) then we can find out the area of the \(\triangle ABC\).can you find out \(AF\) and \(BC\)?

Can you now finish the problem ……….

Second Hint

Area of Triangle

Now clearly $\triangle A D O_{1} \sim \triangle A E O_{2} \sim \triangle A F C$ ( as \(O_1D\) and \(O_2E\) are perpendicular on \(AC\) , R-H-S law )

From Similarity we can say that , $\frac{A O_{1}}{A O_{2}}=\frac{D O_{1}}{E O_{2}} \Rightarrow \frac{A O_{1}}{A O_{1}+3}=\frac{1}{2} \Longrightarrow A O_{1}=3$

By the Pythagorean Theorem we have that $A D=\sqrt{3^{2}-1^{2}}=\sqrt{8}$

Again from $\triangle A D O_{1} \sim \triangle A F C$
$\frac{A D}{A F}=\frac{D O_{1}}{C F} \Longrightarrow \frac{2 \sqrt{2}}{8}=\frac{1}{C F} \Rightarrow C F=2 \sqrt{2}$

can you finish the problem……..

Final Step

The area of the triangle is $\frac{1}{2} \cdot A F \cdot B C=\frac{1}{2} \cdot A F \cdot(2 \cdot C F)=A F \cdot C F=8(2 \sqrt{2})$=\(16\sqrt2\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Problem on Cube | AMC 10A, 2008 | Problem 21

Try this beautiful problem from Geometry: Problem on Cube.

Problem on Cube – AMC-10A, 2008- Problem 21


A cube with side length 1 is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $A B C D ?$

,

 i

  • $\frac{\sqrt{6}}{2}$
  • $\frac{5}{4}$
  • $\sqrt{2}$
  • $\frac{5}{8}$
  • $\frac{3}{4}$

Key Concepts


Geometry

Square

Pythagoras

Check the Answer


But try the problem first…

Answer: $\frac{\sqrt{6}}{2}$

Source
Suggested Reading

AMC-10A (2008) Problem 21

Pre College Mathematics

Try with Hints


First hint

Problem on Cube - figure

The above diagram is a cube and given that side length $1$ and \(B\) and \(D\) are the mid points .we have to find out area of the \(ABCD\).Now since $A B=A D=C B=C D=\sqrt{\frac{1}{2}^{2}}+1^{2},$ it follows that $A B C D$ is a rhombus. can you find out area of the rhombus?

Can you now finish the problem ……….

Second Hint

Problem on Cube - figure

The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$, where $d_1,\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^{2}+1^{2}}=\sqrt{2} \cdot A C$ is a space diagonal, with length $\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$

can you finish the problem……..

Final Step

Therefore area $A=\frac{1}{2} \times \sqrt{2} \times \sqrt{3}=\frac{\sqrt{6}}{2}$

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Television Problem | AMC 10A, 2008 | Problem 14

Try this beautiful Television Problem from AMC – 10A, 2008.

Television Problem – AMC-10A, 2008- Problem 14


Older television screens have an aspect ratio of 4: 3 . That is, the ratio of the width to the height is 4: 3 . The aspect ratio of many movies is not $4: 3,$ so they are sometimes shown on a television screen by “letterboxing” – darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of 2: 1 and is shown on an older television screen with a 27 -inch diagonal. What is the height, in inches, of each darkened strip?

,

 i

Television Problem
  • $2$
  • $2.25$
  • $2.5$
  • $2.7$
  • $3$

Key Concepts


Geometry

Square

Pythagoras

Check the Answer


But try the problem first…

Answer: $2.7$

Source
Suggested Reading

AMC-10A (2008) Problem 14

Pre College Mathematics

Try with Hints


First hint

Television Problem

The above diagram is a diagram of Television set whose aspect ratio of $4: 3$.Suppose a movie has an aspect ratio of $2: 1$ and is shown on an older television screen with a $27$-inch diagonal. Then we have to find the height, in inches, of each darkened strip.

we assume that the width and height of the screen be $4x$ and $3x$ respectively, and let the width and height of the movie be $2y$ and $y$ respectively. If we can find out the value of \(x\) and \(y\) then  the height of each strip can be calculate eassily

Can you now finish the problem ……….

Second Hint

Television Problem

By the Pythagorean Theorem, the diagonal is $\sqrt{(3 x)^{2}+(4 x)^{2}}=5 x=27 .$ So $x=\frac{27}{5}$

Now the movie and the screen have the same width, $2 y=4 x \Rightarrow y=2 x$

can you finish the problem……..

Final Step

Thus, the height of each strip is $\frac{3 x-y}{2}=\frac{3 x-2 x}{2}=\frac{x}{2}=\frac{27}{10}=2.7$

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