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Maximum area | PRMO-2019 | Problem 23

Try this beautiful problem from PRMO, 2019 based on Maximum area

Maximum area | PRMO-2019 | Problem-23

Let $\mathrm{ABCD}$ be a convex cyclic quadrilateral. Suppose $\mathrm{P}$ is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least. If ${\mathrm{PA}, \mathrm{PB}, \mathrm{PC}, \mathrm{PD}}={3,4,6,8} .$ What is the maximum possible area of ABCD?

• $20$
• $55$
• $13$
• $23$

Key Concepts

Geometry

Triangle

Area

Answer:$55$

PRMO-2019, Problem 23

Pre College Mathematics

Try with Hints

Given that $\mathrm{PA}=\mathrm{a}, \mathrm{PB}=\mathrm{b}, \mathrm{PC}=\mathrm{c}, \mathrm{PD}=\mathrm{d}$
Now from the above picture area of quadrilateral ABCD
Area=$[\mathrm{APB}]+[\mathrm{BPC}]+[\mathrm{CPD}]+[\mathrm{DPA}]$

Therefore area $\Delta=\frac{1}{2} \mathrm{ab} \sin \mathrm{x}+\frac{1}{2} \mathrm{bc} \sin \mathrm{y}+\frac{1}{2} \mathrm{cd} \sin \mathrm{z}+\frac{1}{2}$ da $\sin \mathrm{w}$
$\Delta_{\max }$ when $x=y=z=w=90^{\circ}$
$\Delta_{\max }=\frac{1}{2}(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{d})$
Now ac $=$ bd (cyclic quadrilateral) As $(a, b, c, d)=(3,4,6,8)$
$\Rightarrow{(a, c)(b, d)}={(3,8)(4,6)}$

So $\Delta_{\max }=\frac{1}{2} \times 11 \times 10=55$

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Rectangle Pattern | AMC-10A, 2016 | Problem 10

Try this beautiful problem from Geometry based on Rectangle Pattern from AMC 10A, 2016, Problem 10.

Rectangle Pattern- AMC-10A, 2016- Problem 10

A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle?

• $1$
• $2$
• $4$
• $6$
• $8$

Key Concepts

Geometry

Rectangle

square

Answer: $2$

AMC-10A (2016) Problem 10

Pre College Mathematics

Try with Hints

Given that length of the inner rectangle be $x$. Therefore the area of that rectangle is $x \cdot 1=x$
The second largest rectangle has dimensions of $x+2$ and 3 , Therefore area $3 x+6$. Now area of the second shaded rectangle= $3 x+6-x=2 x+6$

can you finish the problem……..

Now the dimension of the largest rectangle is $x+4$ and 5 , and the area= $5 x+20$. The area of the largest shaded region is the largest rectangle- the second largest rectangle, which is $(5 x+20)-(3 x+6)=2 x+14$

can you finish the problem……..

Now The problem states that $x, 2 x+6,2 x+14$ is an arithmetic progression,i.e the common difference will be same . So we can say $(2 x+6)-(x)=(2 x+14)-(2 x+6) \Longrightarrow x+6=8 \Longrightarrow x=2$

Therefore the side length =$2$

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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?

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

Key Concepts

Geometry

Circle

Pythagoras

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

AMC-10A (2009) Problem 21

Pre College Mathematics

Try with Hints

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 ……….

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 ……….

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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Ratio of the areas | PRMO-2019 | Problem 19

Try this beautiful problem from PRMO, 2019 based on Ratio of the areas.

Ratio of the areas | PRMO | Problem-19

Let $\mathrm{AB}$ be a diameter of a circle and let $\mathrm{C}$ be a point on the segment $\mathrm{AB}$ such that $\mathrm{AC}: \mathrm{CB}=6: 7 .$ Let $\mathrm{D}$ be a point on the circle such that $\mathrm{DC}$ is perpendicular to $\mathrm{AB}$. Let DE be the diameter through $\mathrm{D}$. If $[\mathrm{XYZ}]$ denotes the area of the triangle XYZ. Find [ABD] / $[\mathrm{CDE}]$ to the nearest integer.

• $20$
• $91$
• $13$
• $23$

Key Concepts

Geometry

Triangle

Area

Answer:$13$

PRMO-2019, Problem 19

Pre College Mathematics

Try with Hints

$\angle \mathrm{AOC} \quad=\frac{6 \pi}{13}, \angle \mathrm{BOC}=\frac{7 \pi}{13}$

$\mathrm{Ar} \Delta \mathrm{ABD}=\mathrm{Ar} \Delta \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \times \mathrm{OC} \sin \frac{6 \pi}{13}$

$\mathrm{Ar} \Delta \mathrm{CDE}=\frac{1}{2} \mathrm{DE} \times \mathrm{OC} \sin \left(\frac{7 \pi}{13}-\frac{6 \pi}{13}\right)$

$\frac{[\mathrm{ABD}]}{[\mathrm{CDE}]}=\frac{\sin \frac{6 \pi}{13}}{\sin \frac{\pi}{13}}=\frac{1}{2 \sin \frac{\pi}{26}}=\mathrm{p}$

because $\sin \theta \cong \theta$ if $\theta$ is small
$\Rightarrow \sin \frac{\pi}{26} \cong \frac{\pi}{26}$

$\mathrm{p}=\frac{13}{\pi} \Rightarrow$ Nearest integer to $\mathrm{p}$ is 4

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Area of the Inner Square | AMC-10A, 2005 | Problem 8

Try this beautiful problem from Geometry: Area of the inner square

Area of Inner Square – AMC-10A, 2005- Problem 8

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$. What is the area of the inner square $EFGH$?

• $25$
• $32$
• $36$
• $42$
• $40$

Key Concepts

Geometry

Square

similarity

Answer: $36$

AMC-10A (2005) Problem 8

Pre College Mathematics

Try with Hints

We have to find out the area of the region $EFGH$ Which is a square shape .so if we can find out one of it’s side length then we can easily find out the area of $EFGH$. Now given that $BE=1$ i.e $BE=CF=DG=AH=1$ and side length of the square $ABCD=\sqrt {50}$.Therefore $(AB)^2=(\sqrt {50})^2=50$.so using this information can you find out the length of $EH$?

Can you find out the required area…..?

Since $EFGH$ is a square,therefore $ABH$ is a Right -angle Triangle.

Therefore,$(AH)^2+(BH)^2=(AB)^2$

$\Rightarrow (AH)^2+(HE+EB)^2=(AB)^2$

$\Rightarrow (1)^2+(HE+1)^2=50$

$\Rightarrow (HE+1)^2=49$

$\Rightarrow (HE+1)=7$

$\Rightarrow HE=6$

Therefore area of the inner square (red shaded region) =${6}^2=36$

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Triangle and Quadrilateral | AMC-10A, 2005 | Problem 25

Try this beautiful problem from Geometry: Area of Triangle and Quadrilateral

Ratios of the areas of Triangle and Quadrilateral – AMC-10A, 2005- Problem 25

In $ABC$ we have $AB = 25$, $BC = 39$, and $AC=42$. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?

• $\frac{19}{56}$
• $\frac{19}{66}$
• $\frac{17}{56}$
• $\frac{11}{56}$
• $\frac{19}{37}$

Key Concepts

Geometry

Triangle

Answer: $\frac{19}{56}$

AMC-10A (2005) Problem 25

Pre College Mathematics

Try with Hints

Given that $AB = 25$, $BC = 39$, and $AC=42$.we have to find out Ratios of the areas of Triangle$\triangle ADE$ and the quadrilateral $CBED$.So if we can find out the area the $\triangle ADE$ and area of the $\triangle ABC$ ,and subtract $\triangle ADE$ from $\triangle ABC$ then we will get area of the region $CBDE$.Can you find out the area of $CBDE$?

Can you find out the required area…..?

Now $\frac{\triangle ADE}{\triangle ABC}=\frac{AD}{AB}.\frac{AE}{AC}=\frac{19}{25}.\frac{14}{42}=\frac{19}{75}$

Therefore area of $BCED$=area of $\triangle ABC$-area of $\triangle ADE$.Now can you find out Ratios of the areas of Triangle and the quadrilateral?

can you finish the problem……..

Now $\frac{\triangle ADE}{quad.BCED}$=$\frac{\triangle ADE}{{\triangle ABC}-{\triangle ADE}}$=$\frac{1}{\frac{\triangle ABC}{\triangle ADE}-1}=\frac{1}{\frac{75}{19}-1}=\frac{19}{56}$

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Pentagon & Square Pattern | AMC-10A, 2001 | Problem 18

Try this beautiful problem from Geometry based on Pentagon and square Pattern.

Pentagon & Square Pattern – AMC-10A, 2001- Problem 18

The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to

• $50$
• $58$
• $60$
• $56$
• $64$

Key Concepts

Geometry

Pentagon

square

Answer: $56$

AMC-10A (2001) Problem 18

Pre College Mathematics

Try with Hints

The given square is the above square.we have to find out The percent of the plane that is enclosed by the pentagons.Notice that there are $9$ tiles in the square box.so if we can find out the area of pentagon and small square in single tile,then we can find out the total area of the pentagon in the total big square……

can you finish the problem……..

Now consider a single tile from the big square,Let us take the side of the small square is $a$.There are four squares which is in the area $4a^2$ and there are five pentagons which are in areas $5a^2$.Then the area of the single tile is $9a^2$

Therefore we can say that exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane also

can you finish the problem……..

Therefore for the whole square, expressed as a percentage,it becomes $55.\overline{5}\%$, and the closest integer to this value is $56$

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Circular Cylinder Problem | AMC-10A, 2001 | Problem 21

Try this beautiful problem from Geometry based on Circular Cylinder.

Circular Cylinder Problem – AMC-10A, 2001- Problem 21

A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

• $\frac{30}{23}$
• $\frac{30}{11}$
• $\frac{15}{11}$
• $\frac{17}{11}$
• $\frac{3}{2}$

Key Concepts

Geometry

Cylinder

cone

Answer: $\frac{30}{11}$

AMC-10A (2001) Problem 21

Pre College Mathematics

Try with Hints

Given that the diameter equal to its height is inscribed in a right circular cone.Let the diameter and the height of the right circular cone be $2r$.And also The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide.we have to find out the radius of the cylinder.Now if we can show that $\triangle AFE \sim \triangle AGC$, then we can find out the value of $r$

Can you now finish the problem ……….

Given that $Bc=10$,$AG=12$,$HL=FG=2r$. Therefore $AF=12-2r$,$FE=r$,$GC=5$

Now the $\triangle AFE \sim \triangle AGC$, Can you find out the radius from from this similarity property…….?

can you finish the problem……..

Since $\triangle AFE \sim \triangle AGC$, we can write $\frac{AF}{FE}=\frac{AG}{GC}$

$\Rightarrow \frac{12-2r}{r}=\frac{12}{5}$

$\Rightarrow r=\frac{30}{11}$

Therefore the radius of the cylinder is $\frac{30}{11}$

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Area of the Region Problem | AMC-10A, 2007 | Problem 24

Try this beautiful problem from Geometry: Area of the region

Problem on Area of the Region – AMC-10A, 2007- Problem 24

Circle centered at $A$ and $B$ each have radius $2$, as shown. Point $O$ is the midpoint of $\overline{AB}$, and $OA = 2\sqrt {2}$. Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$?

• $\pi$
• $7\sqrt 3 -\pi$
• $8\sqrt 2 -4-\pi$

Key Concepts

Geometry

Triangle

similarity

Answer: $8\sqrt 2 -4-\pi$

AMC-10A (2007) Problem 24

Pre College Mathematics

Try with Hints

We have to find out the area of the region $ECODF$ i.e of gray shaded region.this is not any standard geometrical figure (such as circle,triangle…etc).so we can not find out the value easily.Now if we join $AC$,$AE$,$BD$,$BF$.Then $ABFE$ is a rectangle.then we can find out the required area by [ area of rectangle $ABEF$- (area of arc $AEC$+area of $\triangle ACO$+area of $\triangle BDO$+ area of arc $BFD$)]

Can you find out the required area…..?

Given that Circle centered at $A$ and $B$ each have radius $2$ and Point $O$ is the midpoint of $\overline{AB}$, and $OA = 2\sqrt {2}$

Area of $ABEF$=$2 \times 2 \times 2\sqrt 2$=$8\sqrt 2$

Now $\triangle{ACO}$ is a right triangle. We know $AO=2\sqrt{2}$and $AC=2$, so $\triangle{ACO}$ is isosceles, a $45$-$45$ right triangle.$\overline{CO}$ with length $2$. The area of $\triangle{ACO}=\frac{1}{2} \times base \times height=2$. By symmetry, $\triangle{ACO}\cong\triangle{BDO}$, and so the area of $\triangle{BDO}$ is also $2$.now the $\angle CAO$ = $\angle DBO$=$45^{\circ}$. therefore $\frac{360}{45}=8$

So the area of arc $AEC$ and arc $BFD$=$\frac{1}{8} \times$ area of the circle=$\frac{\pi 2^2}{8}$=$\frac{\pi}{2}$

can you finish the problem……..

Therefore the required area by [ area of rectangle $ABEF$- (area of arc $AEC$+area of $\triangle ACO$+area of $\triangle BDO$+ area of arc $BFD$)]=$8\sqrt 2-(\frac{\pi}{2}+2+2+\frac{\pi}{2}$)=$8\sqrt 2 -4-\pi$

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Problem on Circumscribed Circle | AMC-10A, 2003 | Problem 17

Try this beautiful problem from Geometry based on Circumscribed Circle

Problem on Circumscribed Circle – AMC-10A, 2003- Problem 17

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

• $\frac{5\sqrt3}{\pi}$
• $\frac{3\sqrt3}{\pi}$
• $\frac{3\sqrt3}{2\pi}$

Key Concepts

Geometry

Triangle

Circle

Answer: $\frac{3\sqrt3}{\pi}$

AMC-10A (2003) Problem 17

Pre College Mathematics

Try with Hints

Let ABC is a equilateral triangle which is inscribed in a circle. with center $O$. and also given that perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle.so for find out the peremeter of Triangle we assume that the side length of the triangle be $x$ and the radius of the circle be $r$. then the side of an inscribed equilateral triangle is $r\sqrt{3}$=$x$

Can you now finish the problem ……….

The perimeter of the triangle is=$3x$=$3r\sqrt{3}$ and Area of the circle=$\pi r^2$

Now The perimeter of the triangle=The Area of the circle

Therefore , $3x$=$3r\sqrt{3}$=$\pi r^2$

can you finish the problem……..

Now $3x$=$3r\sqrt{3}$=$\pi r^2$ $\Rightarrow {\pi r}=3\sqrt 3$ $\Rightarrow r=\frac{3\sqrt3}{\pi}$