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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


First hint

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 ……….

Second Hint

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……..

Final Step

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


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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Algebra Math Olympiad PRMO USA Math Olympiad

Circle | Geometry Problem | PRMO-2017 | Question 27

Try this beautiful Problem from Geometry based on Circle from PRMO 2017.

Circle – PRMO 2017, Problem 27


Let $\Omega_{1}$ be a circle with centre 0 and let $A B$ be a diameter of $\Omega_{1} .$ Le $P$ be a point on the segment $O B$ different from 0. Suppose another circle $\Omega_{2}$ with centre P lies in the interior of $\Omega_{1}$. Tangents are drawn from A and B to the circle $\Omega_{2}$ intersecting $\Omega_{1}$ again at $A_{1}$ and $B_{1}$ respectively such that $A_{1}$, and $B_{1}$ are on the opposite sides of $A B$. Given that $A_{1} B=5, A B_{1}=15$ and $O P=10,$ find the radius of $\Omega_{1}$

  • $9$
  • $40$
  • $34$
  • $20$

Key Concepts


Geometry

Circle

Check the Answer


Answer:$20$

PRMO-2017, Problem 27

Pre College Mathematics

Try with Hints


Circle Problem

Let radius of $\Omega_{1}$ be $R$ and that of $\Omega_{2}$ be $r$
From figure, $\Delta \mathrm{ADP} \sim \Delta \mathrm{AA}_{1} \mathrm{B}$
[
\begin{array}{l}
\Rightarrow \frac{D P}{A, B}=\frac{A P}{A B} \
\Rightarrow \frac{r}{5}=\frac{R+10}{2 R}
\end{array}
]

Can you now finish the problem ……….

Circle Problem figure

Again, $\Delta B P E \sim \Delta B A B_{1}$
Therefore $\frac{P E}{A B_{1}}=\frac{B P}{B A}$
$\Rightarrow \frac{r}{15}=\frac{R-10}{2 R}$

Can you finish the problem……..

Dividing (1) by (2)

$3=\frac{R+10}{R-10} \Rightarrow R=20$

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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


Answer: $16 \sqrt{2}$

AMC-10A (2006) Problem 21

Pre College Mathematics

Try with Hints


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 ……….

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……..

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

Circle Problem | AMC 10A, 2006 | Problem 23

Try this beautiful problem from Geometry: Circle

Circle Problem – AMC-10A, 2006- Problem 23


Circles with centers $A$ and $B$ have radii 3 and 8 , respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $A B$ and $C D$ intersect at $E,$ and $A E=5 .$ What is $C D ?$

,

 i

  • $13$
  • $\frac{44}{3} $
  • $\sqrt{221}$
  • $\sqrt{255}$
  • \(\frac{55}{3}\)

Key Concepts


Geometry

Circle

Tangents

Check the Answer


Answer: $ \frac{44}{3}$

AMC-10 (2006) Problem 23

Pre College Mathematics

Try with Hints


Circle Problem

Given that Circles with centers $A$ and $B$ have radii 3 and 8 and $A E=5 .$.we have to find out \(CD\).So join \(BC\) and \(AD\).then clearly \(\triangle BCE\) and \(\triangle ADE\) are Right-Triangle(as \(CD\) is the common tangent ).Now \(\triangle BCE\) and \(\triangle ADE\) are similar.Can you proof \(\triangle BCE\) and \(\triangle ADE\)?

Can you now finish the problem ……….

Circle Problem

$\angle A E D$ and $\angle B E C$ are vertical angles so they are congruent, as are angles $\angle A D E$ and $\angle B C E$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $\triangle A C E \sim \triangle B D E$.

By the Pythagorean Theorem, line segment \(DE=4\)

Therefore from the similarity we can say that \(\frac{D E}{A D}=\frac{C E}{B C} \Rightarrow \frac{4}{3}=\frac{C E}{8}\) .

Therefore \(C E=\frac{32}{3}\)

can you finish the problem……..

Therefore \(CD=CE+DE=4+\frac{32}{3}=\frac{44}{3}\)

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Geometry Math Olympiad PRMO USA Math Olympiad

Finding side of Triangle | PRMO-2014 | Problem 15

Try this beautiful problem from PRMO, 2014 based on Finding side of Triangle.

Finding side of Triangle | PRMO | Problem 15


Let XOY be a triangle with angle XOY=90 degrees. Let M and N be the midpoints of the legs OX and OY, respectively. Suppose that XN=19 and YM=22. what is XY?

  • \(28\)
  • \(26\)
  • \(30\)

Key Concepts


Geometry

Triangle

Pythagoras

Check the Answer


Answer:\(26\)

PRMO-2014, Problem 15

Pre College Mathematics

Try with Hints


Finding side of Triangle - figure

Given that \(\angle XOY=90^{\circ}\) .Let M and N be the midpoints of the legs OX and OY, respectively, and that XN=19 and YM=22. Now \(\triangle XON\) & \(\triangle MOY\) are Right angle Triangle. Use Pythagoras theorem …….

Can you now finish the problem ……….

Triangle problem

Let \(XM=MO=p\) and \(ON=NY=q\).Now using Pythagoras theorm on \(\triangle XON\) & \(\triangle MOY\) we have…

\(OX^2 +ON^2=XN^2\) \(\Rightarrow 4p^2 +q^2=19^2\) \(\Rightarrow 4p^2 +q^2=361\)………..(1) and \(OM^2 +OY^2=MY^2\) \(\Rightarrow p^2 +4q^2=22^2\) \(\Rightarrow p^2 +4q^2=484\)……(2)

Finding side of Triangle

Now Adding (1)+(2)=\((4p^2 +q^2=361)\)+\((p^2 +4q^2=484\) \(\Rightarrow 5(p^2+q^2)=845\) \(\Rightarrow (p^2+q^2)=169\) \(\Rightarrow 4(p^2+q^2)=676\) \(\Rightarrow (OX)^2+(OY)^2=(26)^2\) \(\Rightarrow (XY)^2=(26)^2\) \(\Rightarrow XY=26\).

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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-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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Coordinate Geometry I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB

ISI MStat 2016 PSA Problem 9 | Equation of a circle

This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

Equation of a circle- ISI MStat Year 2016 PSA Question 9


Given \( \theta \) in the range \( 0 \leq \theta<\pi,\) the equation \( 2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0\) represents a circle for all \( \theta\) in the interval

  • \( 0 < \theta <\frac{\pi}{3} \)
  • \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)
  • \( 0 < \theta <\frac{\pi}{2} \)
  • \( 0 \le \theta <\frac{\pi}{2} \)

Key Concepts


Equation of a circle

Trigonometry

Basic Inequality

Check the Answer


Answer: is \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

Try with Hints


Complete the Square.

We get ,

\(2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3)) \)
\(6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2} \)

We are given that \( 0 \leq \theta<\pi,\) . So, \( {\sin^2 \theta} \geq \frac{1}{2} \) \( \Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4} \).

ISI MStat 2016 PSA Problem 9
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Ratio of Circles | AMC-10A, 2009 | Problem 21

Try this beautiful problem from Geometry based on ratio of Circles from AMC 10A, 2009, Problem 21.

Ratio of Circles – AMC-10A, 2009- Problem 21


Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?

Figure of the problem
  • $3-2 \sqrt{2}$
  • $2-\sqrt{2}$
  • $4(3-2 \sqrt{2})$
  • $\frac{1}{2}(3-\sqrt{2})$
  • $2 \sqrt{2}-2$

Key Concepts


Geometry

Circle

Pythagoras

Check the Answer


Answer: \(4(3-2 \sqrt{2})\)

AMC-10A (2009) Problem 21

Pre College Mathematics

Try with Hints


Circles in Circle

We have to find out the ratio of the sum of the areas of the four smaller circles to the area of the larger circle. To find out the area any circle,we need radius.so at first we have to find out radius of two types circle.Can you find out the radius of two type circle i.e small circle and big circle………………

Can you now finish the problem ……….

finding ratio of area of circles

Let the radius of the Four small circles be \(r\).Therfore from the above diagram we can say \(CD=DE=EF=CF=2r\). Now the quadrilateral \(CDEF\) in the center must be a square. Therefore from Pythagoras theorm we can say \(DF=\sqrt{(2r)^2 + (2r)^2}=2r\sqrt 2\). So \(AB=AD+DF+BF=r+2r\sqrt 2+r=2r+2r\sqrt 2\)

Therefore radius of the small circle is \(r\) and big circle is\(R=r+r \sqrt{2}=r(1+\sqrt{2})\)

Can you now finish the problem ……….

shaded circles

Therefore the area of the large circle is \(L=\pi R^{2}=\pi r^{2}(1+\sqrt{2})^{2}=\pi r^{2}(3+2 \sqrt{2}) \)and the The area of four small circles is \(S=4 \pi r^{2}\)

The ratio of the area will be \(\frac{S}{L}=\frac{4 \pi r^{2}}{\pi r^{2}(3+2 \sqrt{2})}\)

=\(\frac{4}{3+2 \sqrt{2}}\)

=\(\frac{4}{3+2 \sqrt{2}} \cdot \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}\)

=\(\frac{4(3-2 \sqrt{2})}{3^{2}-(2 \sqrt{2})^{2}}\)

=\(\frac{4(3-2 \sqrt{2})}{1}\)

=\(4(3-2 \sqrt{2})\)

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