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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

Suggested Book | Source | Answer


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  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

Suggested Book | Source | Answer


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

Suggested Book | Source | Answer


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

Suggested Book | Source | Answer


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

Suggested Book | Source | Answer


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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Geometry Math Olympiad PRMO USA Math Olympiad

Area of the Trapezium | PRMO-2017 | Question 30

Try this beautiful Problem from Geometry based on Area of the Trapezium from PRMO 2017.

Area of the Trapezium – PRMO 2017, Problem 30


Consider the areas of the four triangles obtained by drawing the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium ABCD. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are 1296 and 576 , determine the square root of the maximum possible area of the trapezium to the nearest integer.

  • $9$
  • $40$
  • $13$
  • $20$

Key Concepts


Geometry

Triangle

Trapezium

Check the Answer


Answer:$13$

PRMO-2017, Problem 30

Pre College Mathematics

Try with Hints


Let $x, y, z, w$ be areas of the four triangles as shown in figure.
then area of \(\triangle ADB\)= Area of \(\triangle ACB\)
$\Rightarrow x+y=x+w \Rightarrow y=w$

Also \(\frac{AE}{EC}\)=\(\frac{area of \triangle ADE}{area of \triangle DEF}\)=\(\frac{area of \triangle AEB}{area of \triangle BEC}\)
$\Rightarrow \frac{y}{z}=\frac{x}{w}=\frac{x}{y} \Rightarrow y^{2}=z x$
$\Rightarrow z, y, x$ are in G.P.

Can you now finish the problem ……….

Let $y=z r$ and $x=z r^{2},$ where $r \geq 1$ To make area of trapezium ABCD maximum, we take $z y=z^{2} r=576$
and $y w=z^{2} r^{2}=1296$
As $( z \leq y \leq x)$
Therefore $ \frac{z^{2} r^{2}}{z^{2} r}=\frac{1296}{576} \Rightarrow r=\frac{9}{4} \Rightarrow z=16$

Can you finish the problem……..

Therefore area of trapezium $\mathrm{ABCD}$
$=x+y+z+w=z r^{2}+2 z r+z$
$=z(1+r)^{2}=16\left(1+\frac{9}{4}\right)^{2}=13^{2}$
Therefore Answer is $13 .$

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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


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

AMC-10A (2016) Problem 21

Pre College Mathematics

Try with Hints


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 ……….

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

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

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

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

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

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

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

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

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

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Number of points and planes | AIME I, 1999 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.

Number of points and planes – AIME I, 1999


Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 489
  • is 840
  • cannot be determined from the given information

Key Concepts


Number of points

Plane

Probability

Check the Answer


Answer: is 489.

AIME I, 1999, Question 10

Geometry Vol I to IV by Hall and Stevens

Try with Hints


\(10 \choose 3\) sets of 3 points which form triangles,

fourth distinct segment excluding 3 segments of triangles=45-3=42

Required probability=\(\frac{{10 \choose 3} \times 42}{45 \choose 4}\)

where \({45 \choose 4}\) is choosing 4 segments from 45 segments

=\(\frac{16}{473}\) then m+n=16+473=489.

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Graph Coordinates | AMC 10A, 2015 | Question 12

Try this beautiful Problem on Graph Coordinates from coordinate geometry from AMC 10A, 2015.

Graph Coordinates – AMC-10A, 2015- Problem 12


Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^{2}+x^{4}=2 x^{2} y+1 .$ What is $|a-b| ?$

,

  • $0$
  • $1$
  • $2$
  • $3$
  • \(4\)

Key Concepts


Co-ordinate geometry

graph

Distance Formula

Check the Answer


Answer: $2$

AMC-10A (2015) Problem 12

Pre College Mathematics

Try with Hints


The given points are $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ which are satisfying the equation $y^{2}+x^{4}=2 x^{2} y+1$.

So we can write $y^{2}+\sqrt{\pi}^{4}=2 \sqrt{\pi}^{2} y+1$

Can you now finish the problem ……….

Therefore

$y^{2}+\pi^{2}=2 \pi y+1$
$y^{2}-2 \pi y+\pi^{2}=1$
$(y-\pi)^{2}=1$
$y-\pi=\pm 1$
$y=\pi+1$
$y=\pi-1$

can you finish the problem……..

There are only two solutions to the equation, so one of them is the value of $a$ and the other is $b$. our requirement is $|a-b|$ so between a and b which is greater is not importent…………

So, $|(\pi+1)-(\pi-1)|=2$

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Geometry Math Olympiad USA Math Olympiad

Length and Triangle | AIME I, 1987 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Length and Triangle.

Length and Triangle – AIME I, 1987


Triangle ABC has right angle at B, and contains a point P for which PA=10, PB=6, and \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA. Find PC.

Length and Triangle
  • is 107
  • is 33
  • is 840
  • cannot be determined from the given information

Key Concepts


Angles

Algebra

Triangles

Check the Answer


Answer: is 33.

AIME I, 1987, Question 9

Geometry Vol I to Vol IV by Hall and Stevens

Try with Hints


Let PC be x, \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA=120 (in degrees)

Applying cosine law \(\Delta\)APB, \(\Delta\)BPC, \(\Delta\)CPA with cos120=\(\frac{-1}{2}\) gives

\(AB^{2}\)=36+100+60=196, \(BC^{2}\)=36+\(x^{2}\)+6x, \(CA^{2}\)=100+\(x^{2}\)+10x

By Pathagorus Theorem, \(AB^{2}+BC^{2}=CA^{2}\)

or, \(x^{2}\)+10x+100=\(x^{2}\)+6x+36+196

or, 4x=132

or, x=33.

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