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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

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

AMC-10 (2006) Problem 23

Pre College Mathematics

Try with Hints

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 ……….

$\angle A E D$ 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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Triangle Problem | AMC 10B, 2013 | Problem 16

Try this beautiful problem from Geometry from AMC-10B, 2013, Problem-16, based on triangle

Triangle | AMC-10B, 2013 | Problem 16

In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$?

• $12$
• $13.5$
• $15.5$

Key Concepts

Geometry

Triangle

Pythagorean

Answer:$13.5$

AMC-10B, 2013 problem 16

Pre College Mathematics

Try with Hints

We have to find out the area of AEDC which is divided in four triangles i.e$\triangle APC$,$\triangle APE$, $\triangle PED$, $\triangle CPD$

Now if we find out area of four triangle then we can find out the required area AEDC. Now in the question they supply the information only one triangle i.e $\triangle PED$ such that $PE=1.5$, $PD=2$, and $DE=2.5$ .If we look very carefully the given lengths of the $\triangle PED$ then $( 1.5 )^2 +(2)^2=(2.5)^2$ $\Rightarrow$ $(PE)^2 +(PD)^2=(DE)^2$ i,e $\triangle PED$ is a right angle triangle and $\angle DEP =90^{\circ}$ .so we can easily find out the area of the $\triangle PED$ using formula $\frac{1}{2} \times base \times height$ . To find out the area of other three triangles, we must need the lengths of the sides.can you find out the length of the sides of other three triangles…

Can you now finish the problem ……….

To find out the lengths of the sides of other three triangles:

Given that AD and CD are the medians of the given triangle and they intersects at the point P. .we know the fact that the centroid ($P$) divides each median in a $2:1$ ratio .Therefore AP:PD =2:1 & CP:PE=2:1. Now $PE=1.5$ and $PD=2$ .Therefore AP=4 and CP=3.

And also the angles i.e ($\angle APC ,\angle CPD, \angle APE$) are all right angles as $\angle DPE= 90^{\circ}$

can you finish the problem……..

Area of four triangles :

Area of the $\triangle APC$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times AP \times PC$ = $\frac{1}{2} \times 4 \times 3$ =6

Area of the $\triangle DPC$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times CP \times PD$ = $\frac{1}{2} \times 3 \times 2$ =3

Area of the $\triangle PDE$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times PD \times DE$ = $\frac{1}{2} \times 2 \times 1.5$=1.5

Area of the $\triangle APE$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times AP \times PE$ = $\frac{1}{2} \times 4 \times 1.5 =6$

Total area of ACDE=area of ($\triangle APC$+$\triangle APE$+ $\triangle PED$+ $\triangle CPD$)=$(6+3+1.5+6)=17.5$ sq.unit

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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

Answer:$26$

PRMO-2014, Problem 15

Pre College Mathematics

Try with Hints

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 ……….

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)

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

Try this beautiful problem from Geometry: Area of Triangle and Quadrilateral from AMC 10A, 2005.

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……..

$\frac{[A D E]}{[B C E D]}=\frac{[A D E]}{[A B C]-[A D E]}$
=$\frac{1}{[A B C] /[A D E]-1}$
=$\frac{1}{75 / 19-1}$

=$\frac{19}{56}$

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