Categories

## 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

But try the problem first…

Answer:$13.5$

Source

AMC-10B, 2013 problem 16

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Hint

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……..

Final Step

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

Categories

## 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

But try the problem first…

Answer:$26$

Source

PRMO-2014, Problem 15

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Step

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)

Final Step

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

Categories

## Medians of triangle | PRMO-2018 | Problem 10

Try this beautiful problem from Geometry based on medians of triangle

## Medians of triangle | PRMO | Problem 10

In a triangle ABC, the medians from B to CA is perpendicular to the median from C to AB. If the median from A to BC is 30,determine $(BC^2 +AC^2+AB^2)/100$?

• $56$
• $24$
• $34$

### Key Concepts

Geometry

Medians

Centroid

But try the problem first…

Answer:$24$

Source

PRMO-2018, Problem 10

Pre College Mathematics

## Try with Hints

First hint

We have to find out $(BC^2 +AC^2+AB^2)/100$. So, we have to find out $AB, BC, CA$ at first. Now given that, the medians from B to CA is perpendicular to the median from C to AB and the median from A to BC is 30,

So clearly $\triangle BGC$,$\triangle BGF$,$\triangle EGC$ are right angle triangle.Let $CF=3x$ & $BE =3y$ then clearly $CG=2x$ & $BG= 2y$ given that $AD=30$ SO $AG=20$ & $DG =10$ (as $G$ is centroid, medians intersects at 2:1). Therefore from pythagoras theorem we can find out $BC,BF,CE$ i.e we can find out the value $AB,BC,CA$

Can you now finish the problem ……….

Second Hint

$CE^2=(2x)^2+y^2=4x^2 +y^2$

$BF^2=(2y)^2+x^2=4y^2+x^2$

Also, $CG^2+BG^2=BC^2$ $\Rightarrow 4x^2 + 4y^2={20}^2$ $\Rightarrow x^2+y^2=100$

$AC^2=(2CE)^2=4(4x^2+y^2)$

$AB^2=(2BF)^2=4(4y^2+x^2)$

Can you finish the problem……..

Final Step

$(BC^2 +AC^2+AB^2)=20(x^2+y^2)+20^2=2400$

so, $(BC^2 +AC^2+AB^2)/100$=24

Categories

## Ratio Of Two Triangles | AMC-10A, 2004 | Problem 20

Try this beautiful problem from AMC 10A, 2004 based on Geometry: Ratio Of Two Triangles

## Ratio Of Two Triangles – AMC-10A, 2004- Problem 20

Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$

• $\frac{4}{3}$
• $\frac{3}{2}$
• $\sqrt 3$
• $2$
• $1+\sqrt 3$

### Key Concepts

Square

Triangle

Geometry

But try the problem first…

Answer: $2$

Source

AMC-10A (2002) Problem 20

Pre College Mathematics

## Try with Hints

First hint

We have to find out the ratio of the areas of two Triangles $\triangle DEF$ and $\triangle ABE$.Let us take the side length of $AD$=$1$ & $DE=x$,therefore $AE=1-x$

Now in the $\triangle ABE$ & $\triangle BCF$ ,

$AB=BC$ and $BE=BF$.using Pythagoras theorm we may say that $AE=FC$.Therefore $\triangle ABE \cong \triangle CEF$.So $AE=FC$ $\Rightarrow DE=DF$.Therefore the $\triangle DEF$ is  an isosceles right triangle. Can you find out the area of isosceles right triangle $\triangle DEF$

Can you now finish the problem ……….

Second Hint

Length of $DE=DF=x$.Then the the side length of $EF=X \sqrt 2$

Therefore the area of $\triangle DEF= \frac{1}{2} \times x \times x=\frac{x^2}{2}$ and area of $\triangle ABE$=$\frac{1}{2} \times 1 \times (1-x) = \frac{1-x}{2}$.Now from the Pythagoras theorm $(1-x)^2 +1 =2x^2 \Rightarrow x^2=2-2x=2(1-x)$

can you finish the problem……..

Final Step

The ratio of the area of $\triangle DEF$ to that of $\triangle ABE$ is $\frac{\frac{x^2}{2}}{\frac{(1-x)}{2}}$=$\frac{x^2}{1-x}$=$\frac {2(1-x)}{(1-x)}=2$

Categories

## Length of a Tangent | AMC-10A, 2004 | Problem 22

Try this beautiful problem from American Mathematics Competitions, AMC 10A, 2004 based on Geometry: Length of a Tangent.

## Length of a Tangent – AMC-10A, 2004- Problem 22

Square $ABCD$ has a side length $2$. A Semicircle with diameter $AB$ is constructed inside the square, and the tangent  to the semicircle from $C$ intersects side $AD$at $E$. What is the length of $CE$?

• $\frac{4}{3}$
• $\frac{3}{2}$
• $\sqrt 3$
• $\frac{5}{2}$
• $1+\sqrt 3$

### Key Concepts

Square

Semi-circle

Geometry

But try the problem first…

Answer: $\frac{5}{2}$

Source

AMC-10A (2004) Problem 22

Pre College Mathematics

## Try with Hints

First hint

We have to find out length of $CE$.Now $CE$ is a tangent of inscribed the semi circle .Given that length of the side is $2$.Let $AE=x$.Therefore $DE=2-x$. Now $CE$ is the tangent of the semi-circle.Can you find out the length of $CE$?

Can you now finish the problem ……….

Second Hint

Since $EC$ is tangent,$\triangle COF$ $\cong$ $\triangle BOC$ and $\triangle EOF$ $\cong$ $\triangle AOE$ (By R-H-S law).Therefore $FC=2$ & $EC=x$.Can you find out the length of $EC$?

can you finish the problem……..

Final Step

Now the $\triangle EDC$ is a Right-angle triangle……..

Therefore $ED^2+ DC^2=EC^2$ $\Rightarrow (2-x)^2 + 2^2=(2+x)^2$ $\Rightarrow x=\frac{1}{2}$

Hence $EC=EF+FC=2+\frac{1}{2}=\frac{5}{2}$

Categories

## Problem based on Triangles | PRMO-2018 | Problem 12

Try this beautiful problem from PRMO, 2018 based on Triangles.

## Problem based on Triangles | PRMO | Problem 12

In the triangle, ABC, the right angle at A. The altitude through A and the internal bisector of $\angle A$ have lengths $3$ and $4$, respectively. Find the length of the medium through A?

• $24$
• $30$
• $22$
• $18$

### Key Concepts

Geometry

Triangle

Pythagoras

But try the problem first…

Answer:$24$

Source

PRMO-2018, Problem 12

Pre College Mathematics

## Try with Hints

First hint

We have to find out the value of $AM$ .Now we can find out the area of Triangle dividing two parts , area of $\triangle AMC$ + area of $\triangle ABM$ ( as M is the mid point of BC)

Can you now finish the problem ……….

Second Step

Now $AN$ is the internal bisector…Therefore $\angle NAB=\angle NAC= 45^{\circ}$.Let $AC=b$ ,$AB=c$ and $BC=c$.using this values find out the area of triangles AMC and Triangle ABM

Final Step

Area of $\triangle ABC$=$\frac{1}{2} bc=\frac{1}{2} \times a \times 3$

$\Rightarrow bc=3a$………………….(1)

Now Area of $\triangle ABN$ + Area of $\triangle ANC$=Area of $\triangle ABC$

$\Rightarrow \frac{1}{2} c 4 sin 45^{\circ} +\frac{1}{2} b 4 sin 45^{\circ}=\frac{1}{2} bc$

$\Rightarrow b+c =\frac{1}{2\sqrt 2} bc$

(squarring both sides we get……….)

$\Rightarrow b^2 +c^2 +2bc=\frac{1}{8} b^2 c^2$

$\Rightarrow a^2 +6a=\frac{9}{8} a^2$ (from 1)

$\Rightarrow a +6 =\frac{9}{8} a$ $(as a \neq 0)$

$\Rightarrow a=48$

$\Rightarrow AM=BM=MC=\frac{a}{2}=24$

Categories

## Problem based on Triangle | PRMO-2012| Problem 7

Try this beautiful problem from PRMO, 2012 based on Triangle.

## Triangle | PRMO | Problem 12

In $\triangle ABC$ we have $AC=BC=7$ and $AB=2$.Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ .what is the length of the segment $DB$?

• $5$
• $3$
• $7$

### Key Concepts

Geometry

Triangle

Pythagoras

But try the problem first…

Answer:$3$

Source

PRMO-2012, Problem 7

Pre College Mathematics

## Try with Hints

First hint

Given that $AC=BC=7$ & $CD=8$.we have to find out $BD$.Let $BD=x$.we draw a perpendicular from $C$ to $AB$ at the point $M$.Therefore clearly $\triangle CMB$ & $\triangle CMD$ are right angle.so if we can find out the value of $CM$ from the $\triangle CMB$ then we can find out the value $BD$ from the $\triangle CMD$

Can you now finish the problem ……….

Second Step

From the above diagram,In $\triangle CMB$ we can say that $CM=\sqrt{49-1}=4\sqrt 3$

Given $AB=2$ and $M$ is the mid point of $\triangle ABC$ (As AC=BC=7,Isosceles triangle),

Therefore $BM=1$, So $MD=x+1$

Final Step

From the $\triangle CMD$, $(X+1)^2+(4\sqrt 3)^2=64$ $\Rightarrow x=3,-5$

we will take the positive value ,so $BD=3$

Categories

## Problem based on Triangle | PRMO-2016 | Problem 10

Try this beautiful problem from PRMO, 2016 based on Triangle

## Triangle | PRMO | Problem 10

In a triangle ABC right angled at vertex B, a point O is choosen on the side BC such that the circle Y centerer at O of radius OB touches the side AC.Let AB =63 and BC=16 and the radius of Y be the form mn wherw m,n are relatively prime positive integers,find the value of m+n?

• $75$
• $80$
• $71$

### Key Concepts

Geometry

Triangle

Circle

But try the problem first…

Answer:$71$

Source

PRMO-2016, Problem 10

Pre College Mathematics

## Try with Hints

First hint

Given that ABC is a triangle where AB=63 and BC=16 and $\angle ABC=90^{\circ}$. O is any point on BC.Let OB=$r$=radius of the circle.we have to find out the value of $r$ for $m+n$.

Now $\triangle ABC$ is aright angle triangle at$B$ So using pythagoras theorm $AC=\sqrt {(AB)^2+(BC)^2}=\sqrt{(63)^2+(16)^2}=65$.now $\triangle ODC$ is also a right angle triangle as OD radius and AC tangent,can you find out the value of $r$?

Can you now finish the problem ……….

Second Step

Given bc=16,Let OB=$r$ then OC=$(16- r)$.we have to find out the value of $r$.if we can show that $\triangle ABC \sim \triangle ODC$ then we can find out the value of r.can you proof $\triangle ABC \sim \triangle ODC$?

Second Step

In $\triangle ABC$ & $\triangle ODC$,

$\angle ABC=\angle ODC$

$\angle C= \angle C$

Therefore $\triangle ABC \sim \triangle ODC$

so we have ….

$\frac{AC}{OC}=\frac{AB}{OD}=\frac{BC}{AC}$ $\Rightarrow \frac{65}{16-r}=\frac{63}{r}=\frac{16}{65}$

Therefore,$\frac{65}{16-r}=\frac{63}{r}\Rightarrow r=\frac{63}{8}$=$\frac{m}{n}$

Therefore $(m+n)=63+8=71$

Categories

## Radius of a Semi Circle -AMC 8, 2017 – Problem 22

Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.

## AMC-8(2017) – Geometry (Problem 22)

In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?

• $\frac{7}{6}$
• $\frac{10}{3}$
• $\frac{9}{8}$

### Key Concepts

Geometry

congruency

similarity

But try the problem first…

Answer:$\frac{10}{3}$

Source

AMC-8(2017)

Pre College Mathematics

## Try with Hints

First hint

Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.

Can you now finish the problem ……….

Second Hint

Now the $\triangle ODB$and $\triangle OCB$ are congruent

can you finish the problem……..

Final Step

Let x be the radius of the semi circle

Now the $\triangle ODB$ and $\triangle OCB$ we have

OD=OC

OB=OB

$\angle ODB$=$\angle OCB$= 90 degree`

so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)

BD=BC=5

And also $\triangle ODA$ and $\triangle BCA$ are similar….

$\frac{8}{12}$=$\frac{x}{5}$

i.e x =$\frac{10}{3}$

Categories

## Radius of a Circle – SMO 2013 – Problem 25

Try this beautiful problem from Geometry based on the radius and tangent of a circle.

## SMO 2013 – Geometry (Problem 25)

As shown in the figure below ,circles $C_1$and$C_2$ of radius 360 are tangent to each other , and both tangent to the straight line l.if the circle$C_3$ is tangent to $C_1$ ,$C_2$ and l ,and circle$C_4$is tangent to$C_1$,$C_3$ and l ,find the radius of$C_4$

• 30
• 35
• 40

### Key Concepts

Geometry

Pythagoras theorm

Distance Formula

But try the problem first…

Source

Pre College Mathematics

## Try with Hints

First hint

Let R be the radius of $C_3$

$C_2E$ =360-R

$C_3E=360$

$C_2C_3$=360+R

Using pythagoras theorm ….

$(360-R)^2+360^2=(360+R)^2$

i.e R=90

Can you now finish the problem ……….

Second Hint

Let the radius of$C_4$ be r

then use the distacce formula and tangent property……..

can you finish the problem……..

Final Step

Let r be the radius of $C_4$ (small triangle).

LO+OC=360

$\sqrt{(360+p)^2-(360-p)^2}+\sqrt{(90+r)^2-(90-r)^2}=360$

i.e r=40.