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AMC 8 Math Olympiad USA Math Olympiad

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 \(\triangle ABE\)

Ratio of two triangles - problem
  • \(\frac{4}{3}\)
  • \(\frac{3}{2}\)
  • \(\sqrt 3\)
  • \(2\)
  • \(1+\sqrt 3\)

Key Concepts


Square

Triangle

Geometry

Check the Answer


Answer: \(2\)

AMC-10A (2002) Problem 20

Pre College Mathematics

Try with Hints


Shaded triangles

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 ……….

Shaded triangular regions

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……..

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

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AMC 8 Math Olympiad USA Math Olympiad

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 \(CE\)?

Length of a Tangent - Problem
  • \(\frac{4}{3}\)
  • \(\frac{3}{2}\)
  • \(\sqrt 3\)
  • \(\frac{5}{2}\)
  • \(1+\sqrt 3\)

Key Concepts


Square

Semi-circle

Geometry

Check the Answer


Answer: \(\frac{5}{2}\)

AMC-10A (2004) Problem 22

Pre College Mathematics

Try with Hints


Length of a Tangent - Problem figure

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 ……….

Shaded figure 2
Shaded figure 1

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……..

Shaded Triangle to find the length of the tangent

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 \(EC=EF+FC=2+\frac{1}{2}=\frac{5}{2}\)

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AMC 8 Math Olympiad USA Math Olympiad

Diamond Pattern | AMC-10A, 2009 | Problem 15

Try this beautiful problem from AMC 10A, 2009 based on Diamond Pattern.

Diamond Pattern – AMC-10A, 2009- Problem 15


The figures \(F_1\), \(F_2\), \(F_3\), and \(F_4\) shown are the first in a sequence of figures. For \(n\ge3\), \(F_n\) is constructed from \(F_{n – 1}\) by surrounding it with a square and placing one more diamond on each side of the new square than \(F_{n – 1}\) had on each side of its outside square. For example, figure \(F_3\) has \(13\) diamonds. How many diamonds are there in figure \(F_{20}\)?

Diamond Pattern
  • \(756\)
  • \(761\)
  • \(786\)

Key Concepts


Pattern

Sequence

Symmetry

Check the Answer


Answer: \(761\)

AMC-10A (2009) Problem 15

Pre College Mathematics

Try with Hints


Diamond Pattern

From the above diagram we observe that in \(F_1\) the number of diamond is \(1\).in \(F_2\) the number of diamonds are \(5\).in \(F_3\) the number of diamonds are \(13\). in \(F_4\) the numbers of diamonds are \(25\).Therefore from \(F_1\) to \(F_2\) ,\((5-1)\)=\(4\) new diamonds added.from \(F_2\) to \(F_3\),\((13-5=8\) new diamonds added.from \(F_3\) to \(F_4\),\((25-13)=12\) new diamonds added.we may say that When constructing \(F_n\) from \(F_{n-1}\), we add \(4(n-1)\) new diamonds.

Can you now finish the problem ……….

so we may construct that Let \(S_n\) be the number of diamonds in \(F_n\). We already know that \(P_1\)=1 and for all \(n >1\) ,\(P_n=P_{n-1}+4(n-1)\).now can you find out \(P_{20}\)

can you finish the problem……..

Now \(P_{20}\)=\(P_{19} + 4(20-1)\)

\(\Rightarrow P_{20}\)=\(P_{19} + 4.19)\)

\(\Rightarrow P_{20}\)=\(P_{18}+(4 \times 18) +( 4 \times 19)\)

\(\Rightarrow P_{20}\)= …………………………………..

\(\Rightarrow P_{20}\)=\(1+4(1+2+3+……..+18+19)\)

\(\Rightarrow P_{20}\)=\(1+ \frac{ 4 \times 19 \times 20}{2}\)

\(\Rightarrow P_{20}\)=\(761\)

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AMC 8 Math Olympiad USA Math Olympiad

Problem on Semicircle | AMC 8, 2013 | Problem 20

Try this beautiful problem from Geometry based on Semicircle.

Area of the Semicircle – AMC 8, 2013 – Problem 20


A $1\times 2$ rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?

  • \( \frac{\pi}{2}\)
  • \(\pi\)
  • \( \frac{\pi}{3}\)

Key Concepts


Geometry

Square

semi circle

Check the Answer


Answer:\(\pi\)

AMC-8 (2013) Problem 20

Pre College Mathematics

Try with Hints


Problem on Semicircle

At first we have to find out the radius of the semicircle for the area of the semicircle.Now in the diagram,AC is the radius of the semicircle and also AC is the hypotenuse of the right Triangle ABC.

Can you now finish the problem ……….

Now AB=1 AND AC=1 (As ABDE $1\times 2$ rectangle .So using pythagorean theorm we can eassily get the value of AC .and area of semicircle =\(\frac{\pi r^2}{2}\)

can you finish the problem……..

Problem on Semicircle

Given that ABDE is a square whose AB=1 and BD=2

Therefore BC=1

Clearly AC be the radius of the given semi circle

From the \(\triangle ABC\),\((AB)^2+(BC)^2=(AC)^2\Rightarrow AC=\sqrt{(1^2+1^2)}=\sqrt2\)

Therefore the area of the semicircle=\(\frac{1}{2}\times \pi(\sqrt 2)^2\)=\(\pi\)

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AMC 8 Math Olympiad USA Math Olympiad

Radius of semicircle | AMC-8, 2013 | Problem 23

Try this beautiful problem from Geometry: Radius of the semicircle

Radius of the semicircle- AMC-8, 2013- Problem 23


Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$, and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$. What is the radius of the semicircle on $\overline{BC}$?

radius of semicircle
  • \(9\)
  • \(7.5\)
  • \(6\)

Key Concepts


Geometry

Triangle

Semi-circle

Check the Answer


Answer: \(7.5\)

AMC-8 (2013) Problem 23

Pre College Mathematics

Try with Hints


We have to find out the radius of the semi-circle on ${BC}$? Now ABC is a Right angle Triangle.so if you find out AB and AC then BC will be easily calculated…..

Can you now finish the problem ……….

To find the value of AB and AC, notice that area of the semi-circle on AB is given and length of the arc of AC is given….

can you finish the problem……..

radius of semicircle

Let the length of AB=\(2x\),So the radius of semi-circle on AB=\(x\).Therefore the area =\(\frac{1}{2}\times \pi (\frac{x}{2})^2=8\pi\)\(\Rightarrow x=4\)

Theregfore length of AB=\(2x\)=8

Given that the arc of the semi-circle on $\overline{AC}$ has length $8.5\pi$,let us take the radius of the semicircle on AB =\(r\).now length of the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$ i.e half perimeter=\(8.5\)

so \(\frac{1}{2}\times {2\pi r}=8.5\pi\)\(\Rightarrow r=8.5\)\(\Rightarrow 2r=17\).so AC=17

The triangle ABC is a Right Triangle,using pythagorean theorm……

\((AB)^2 + (BC)^2=(AC)^2\)\(\Rightarrow BC=\sqrt{(AC)^2 -(AB)^2}\)\(\Rightarrow BC =\sqrt{(17)^2 -(8)^2}\)\(\Rightarrow BC= \sqrt{289 -64 }\)\(\Rightarrow BC =\sqrt {225}=15\)

Therefore the radius of semicircle on BC =\(\frac{15}{2}=7.5\)

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AMC 8 Math Olympiad USA Math Olympiad

Area of a Regular Hexagon | AMC-8, 2012 | Problem 23

Try this beautiful problem from Geometry: Area of the Regular Hexagon – AMC-8, 2012 – Problem 23.

Area of the Regular Hexagon – AMC-8, 2012- Problem 23


An equilateral triangle and a regular hexagon have equal perimeters. If the triangle’s area is 4, what is the area of the hexagon?

  • \(8\)
  • \(6\)
  • \(10\)

Key Concepts


Geometry

Triangle

Hexagon

Check the Answer


Answer: \(6\)

AMC-8 (2012) Problem 23

Pre College Mathematics

Try with Hints


To find out the area of the Regular hexagon,we have to find out the side length of it.Now the perimeter of the triangle and Regular Hexagon are same….from this condition you can easily find out the side length of the regular Hexagon

Can you now finish the problem ……….

Let the side length of an equilateral triangle is\(x\).so the perimeter will be \(3x\) .Now according to the problem the perimeter of the equiliteral triangle and regular hexagon are same,i.e the perimeter of regular hexagon=\(3x\)

So the side length of be \(\frac{3x}{6}=\frac{x}{2}\)

can you finish the problem……..

Now area of the triangle \(\frac{\sqrt 3}{4}x^2=4\)

Now the area of the Regular Hexagon=\(\frac{3\sqrt3}{2} (\frac{x}{2})^2=\frac{3}{2} \times \frac{\sqrt{3}}{4}x^2=\frac{3}{2} \times 4\)=6

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of a Triangle | AMC-8, 2000 | Problem 25

Try this beautiful problem from Geometry: Area of a Triangle

Area of the Triangle- AMC-8, 2000- Problem 25


The area of rectangle \(ABCD\) is \(72\) units squared. If point \(A\) and the midpoints of \(BC\) and \(CD\) are joined to form a triangle, the area of that triangle is

Area of a Triangle
  • \(25\)
  • \(27\)
  • \(29\)

Key Concepts


Geometry

Triangle

square

Check the Answer


Answer: \(27\)

AMC-8 (2000) Problem 25

Pre College Mathematics

Try with Hints


Area of the triangle =\(\frac{1}{2} \times base \times height \)

Can you now finish the problem ……….

Area of Shaded Triangle

Therefore area of the shaded region i.e area of the \(\triangle AEF\)=area of the square- area of \((\triangle ADE +\triangle EFC +\triangle ABF)\)

can you finish the problem……..

Area of Shaded Triangle

Given that area of the rectangle ABCD=72

Let length AB=\(x\) and length of CD=\(y\)

Therefore DE=EC=\(\frac{y}{2}\) and BF=FC=\(\frac{x}{2}\)

Area of ABCD=\(xy\)=72

Area of the \(\triangle ADE=\frac{1}{2}\times DE \times AD= \frac{1}{2}\times \frac{x}{2} \times y =\frac{xy}{4}=\frac{72}{4}=18\)

Area of the \(\triangle EFC=\frac{1}{2}\times EC \times FC= \frac{1}{2}\times \frac{x}{2} \times \frac{y}{2} =\frac{xy}{8}=\frac{72}{8}=9\)

Area of the \(\triangle ABF=\frac{1}{2}\times AB \times BF= \frac{1}{2}\times y\times \frac{x}{2}=\frac{xy}{4}=\frac{72}{4}=18\)

Therefore area of the shaded region i.e area of the \(\triangle AEF\)=area of the square- area of \((\triangle ADE +\triangle EFC +\triangle ABF)=72-(18+8+18)=27\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of the Trapezoid | AMC 8, 2002 | Problem 20

Try this beautiful problem from Geometry based on Area of Trapezoid.

Area of the Trapezoid – AMC- 8, 2002 – Problem 20


The area of triangle XYZ is 8  square inches. Points  A and B  are midpoints of congruent segments  XY and XZ . Altitude XC bisects YZ.What is the area (in square inches) of the shaded region?

aera of trapezoid
  • \( 6\)
  • \( 4\)
  • \( 3\)

Key Concepts


Geometry

Triangle

Trapezoid

Check the Answer


Answer:\(3\)

AMC-8 (2002) Problem 20

Pre College Mathematics

Try with Hints


Given that Points  A and B  are midpoints of congruent segments  XY and XZ and Altitude XC bisects YZ

Let us assume that the length of YZ=\(x\) and length of \(XC\)= \(y\)

Can you now finish the problem ……….

Therefore area of the trapezoid= \(\frac{1}{2} \times (YC+AO) \times OC\)

can you finish the problem……..

Triangle with centre O

Let us assume that the length of YZ=\(x\) and length of \(XC\)= \(y\)

Given that area of \(\triangle xyz\)=8

Therefore \(\frac{1}{2} \times YZ \times XC\)=8

\(\Rightarrow \frac{1}{2} \times x \times y\) =8

\(\Rightarrow xy=16\)

Given that Points  A and B  are midpoints of congruent segments  XY and XZ and Altitude XC bisects YZ

Then by the mid point theorm we can say that \(AO=\frac{1}{2} YC =\frac{1}{4} YZ =\frac{x}{4}\) and \(OC=\frac{1}{2} XC=\frac{y}{2}\)

Therefore area of the trapezoid shaded area = \(\frac{1}{2} \times (YC+AO) \times OC\)= \(\frac{1}{2} \times ( \frac{x}{2} + \frac{x}{4} ) \times \frac{y}{2}\) =\(\frac{3xy}{16}=3\) (as \(xy\)=16)

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AMC 8 Math Olympiad

Intersection of two Squares | AMC 8, 2004 | Problem 25

Try this beautiful problem from Geometry based on Intersection of two Squares.

When 2 Squares intersect | AMC-8, 2004 | Problem 25


Two \(4\times 4\) squares intersect at right angles, bisecting their intersecting sides, as shown. The circle’s diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

Intersection of two Squares
  • \(28-2\pi\)
  • \(25-2\pi\)
  • \(30-2\pi\)

Key Concepts


Geometry

square

Circle

Check the Answer


Answer: \(28-2\pi\)

AMC-8, 2004 problem 25

Pre College Mathematics

Try with Hints


Area of the square is \(\pi (r)^2\),where \(r\)=radius of the circle

Can you now finish the problem ……….

Clearly, if 2 squares intersect, it would be a smaller square with half the side length, 2.

can you finish the problem……..

Intersection of two Squares

Clearly the intersection of 2 squares would be a smaller square with half the side length, 2.

The area of this region =Total area of larger two squares – the area of the intersection, the smaller square i.e \(4^2+4^2 -2^2=28 \)

Now The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle

Using the Pythagorean th. diameter of the circle be \(\sqrt{2^2 +2^2}=2\sqrt 2\)

Radius=\(\sqrt 2\)

area of the square=\(\pi (\sqrt2)^2\)=\(2\pi\)

Area of the shaded region= 28-2\(\pi\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Radius of the Circle | AMC-8, 2005 | Problem 25

Try this beautiful problem from Geometry: Radius of a circle

Radius of a circle – AMC-8, 2005- Problem 25


A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?

radius of the circle
  • $\frac{5}{\sqrt \pi}$
  • $ \frac{2}{\sqrt \pi} $
  • $\sqrt \pi$

Key Concepts


Geometry

Cube

square

Check the Answer


Answer: $ \frac{2}{\sqrt \pi} $

AMC-8 (2005) Problem 25

Pre College Mathematics

Try with Hints


The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Can you now finish the problem ……….

Shaded region of the figure

Region within the circle and square be \(x\) i.e In other words, it is the area inside the circle and the square

The area of the circle -x=Area of the square – x

can you finish the problem……..

Shaded region of the figure

Given that The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Let the region within the circle and square be \(x\) i.e In other words, it is the area inside the circle and the square .

Let r be the radius of the circle

Therefore, The area of the circle -x=Area of the square – x

so, \(\pi r^2 – x=4-x\)

\(\Rightarrow \pi r^2=4\)

\(\Rightarrow r^2 = \frac{4}{\pi}\)

\(\Rightarrow r=\frac{2}{\sqrt \pi}\)

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