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AMC 10 Math Olympiad USA Math Olympiad

Area of Hexagon Problem | AMC-10A, 2014 | Problem 13

Try this beautiful problem from Geometry based on Area of Hexagon Problem

Area of Hexagon Problem – AMC-10A, 2014- Problem 13


Equilateral $\triangle ABC$ has side length $1$, and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$?

area of hexagon

  • \(3+{\sqrt 5}\)
  • \(4+{\sqrt 3}\)
  • \(3+{\sqrt 3}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{9}\)

Key Concepts


Geometry

Triangle

square

Check the Answer


Answer: \(3+{\sqrt 3}\)

AMC-10A (2014) Problem 13

Pre College Mathematics

Try with Hints


shaded hexagon

Given that \(\triangle ABC\) is an Equilateral Triangle with side length \(1\) and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle.Now we have to find out the area of hexagon $DEFGHI$.Now area of the the Hexagon $DEFGHI$=Area of \(\triangle ABC\)+Area of three squares( i.e $ABDE$, $BCHI$, $CAFG$)+Area of Triangles(( \(\triangle AEF,\triangle DBI,\triangle HCG\))

Since the side length of Equilateral Triangle \(\triangle ABC\) is given then we can find out the area of the \(\triangle ABC\) and area of the squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle(as side length of one square =side length of the equilateral \(\triangle ABC\).Now we have to find out the area of other three Triangles( \(\triangle AEF,\triangle DBI,\triangle HCG\))

can you finish the problem……..

Area of the \(\triangle ABC\)(Red shaded Region)=\(\frac{\sqrt 3}{4}\) (as side lengtjh is 1)

Area of 3 squares =\(3\times {1}^2=3\)

solution figure of hexagon problem

Now we have to find out the area of the \(\triangle GCH\).At first draw a perpendicular \(CL\) on \(HG\). As \(\triangle GCH\) is an isosceles triangle (as \(HC=CG=1\)),Therefore \(HL=GL\)

Now in the \(\triangle CGL\),

\(\angle GCL=60^{\circ}\) (as \(\angle GCH=360^{\circ}-\angle ACB -\angle ACG-\angle BCH \) \(\Rightarrow \angle GCH=360^{\circ}-60^{\circ}-90^{\circ}-90^{\circ}=120^{\circ}\))

So \(\angle GCL=60^{\circ}\)

So \(\angle CGL=30^{\circ}\)

\(\frac{CL}{CG}\)=Sin \(30^{\circ}\)

\(\Rightarrow CL=\frac{1}{2}\) (as CG=1)

And ,

\(\frac{GL}{CG}\)=Sin \(60^{\circ}\)

\(\Rightarrow GL=\frac{\sqrt 3}{2}\) (as CG=1)

So \(GH=\sqrt 3\)

Therefore area of the \(\triangle CGH=\frac{1}{2}\times \sqrt 3 \times{1}{2}=\frac{\sqrt 3}{4}\)

Therefore area of three Triangles ( \(\triangle AEF,\triangle DBI,\triangle HCG\))=\(3\times \frac{\sqrt 3}{4}\)

can you finish the problem……..

Shaded area of hexagon

Therefore area of the the Hexagon $DEFGHI$=Area of \(\triangle ABC\)+Area of three squares( i.e $ABDE$, $BCHI$, $CAFG$)+Area of Triangles(( \(\triangle AEF,\triangle DBI,\triangle HCG\))=(\(\frac{\sqrt 3}{4}+3+3\times \frac{\sqrt 3}{4}\))=\(3+{\sqrt 3}\)

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AMC 10 Math Olympiad USA Math Olympiad

Area of Region in a Circle | AMC-10A, 2011 | Problem 18

Try this beautiful problem from Geometry based on Area of Region in a Circle.

Area of Region in a Circle – AMC-10A, 2011- Problem 18


Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$. What is the area inside Circle $C$ but outside circle $A$ and circle $B$ ?

  • \(\pi\)
  • \(\frac{3\pi}{2}\)
  • \(2\)
  • \(6\)
  • \(\frac{5\pi}{2}\)

Key Concepts


Geometry

Circle

Rectangle

Check the Answer


Answer: \(2\)

AMC-10A (2011) Problem 18

Pre College Mathematics

Try with Hints


We have to find out the area of the shaded region .Given that three circles with radius \(1\) and Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$.so if we draw a rectangle as shown in given below then we can find out the required region by the area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$

can you finish the problem……..

Now area of the rectangle is \(2\times 1=2\)

Area of the half circle with center (gray shaded region)=\(\frac{\pi (1)^2}{2}\)

The area of the two sectors created by $A$ and $B$(blue region)=\(\frac{2\pi(1)^2}{4}\)

can you finish the problem……..

Therefore, the required region (gray region)=area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$=\(\frac{\pi (1)^2}{2}\)+\(2\times 1=2\)-\(\frac{2\pi(1)^2}{4}\)=\(2\)

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AMC 10 Math Olympiad USA Math Olympiad

Sectors in Circle | AMC-10A, 2012 | Problem 10

Try this beautiful problem from Geometry based on Sectors in Circle.

Sectors in Circle – AMC-10A, 2012- Problem 10


Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?

  • \(6\)
  • \(12\)
  • \(14\)
  • \(8\)
  • \(16\)

Key Concepts


Geometry

Circle

AP

Check the Answer


Answer: \(8\)

AMC-10A (2012) Problem 10

Pre College Mathematics

Try with Hints


We have to find out  the degree measure of the smallest possible sector angle.Let $x$ be the smallest sector angle and $r$ be the difference between consecutive sector angles,

Therefore the angles are $x, x+r, a+2r, \cdots. x+11r$. Now we know that sum of the angles of all sectors of a circle is \(360^{\circ}\).Can you find out the values of \(x\) and \(r\)?

can you finish the problem……..

Therefore using the AP formula we will get ,

\(\frac{x+x+11r}{2} . 12=360\)

\(\Rightarrow x=\frac{60-11r}{2}\)

can you finish the problem……..

Since all sector angles are integers so $r$ must be a multiple of 2. Now an even integers for $r$ starting from 2 to minimize $x.$ We find this value to be 4 and the minimum value of $x$ to be \(\frac{60-11(4)}{2}=8\)

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AMC 10 Math Olympiad USA Math Olympiad

Area of quadrilateral | AMC-10A, 2020 | Problem 20

Try this beautiful problem from Geometry based on the Area of the quadrilateral.

Area of the quadrilateral – AMC-10A, 2020- Problem 20


Quadrilateral \(ABCD\) satifies \(\angle ABC= \angle ACD=90^{\circ}\), \(AC=20\) and \(CD=30\). Diagonals \(AC\) and \(BD\) intersect at the Point \(E\) and \(AE=5\). What is the area quadrilateral \(ABCD\)?

  • \(330\)
  • \(340\)
  • \(350\)
  • \(360\)
  • \(370\)

Key Concepts


Geometry

Triangle

Area

Check the Answer


Answer: \(360\)

AMC-10A (2020) Problem 20

Pre College Mathematics

Try with Hints


Area of quadrilateral

Given that \(AC=20\) and \(Cd=30\).Area of \(\triangle ACD=300\).Now if we have to find out the area of \(\triangle ABC\) then we can find out the area of whole \(ABCD\).Now draw altitude from $B$ to $AC$ and call the point of intersection $F$.Let \(FE=x\) .Noe \(AE=5\) then \(AF=5-X\).Now observe that the \(\triangle BEF \sim \triangle DCE\).

Can you now finish the problem ……….



Now $EC$ is $20-5=15$, and $DC=30$, we get that $BF=2x$. Now, if we drawn another diagram $ABC$, we get that $(2x)^2=(5-x)(15+x)$ as we know that the altitude geometric mean theorem which states that the altitude squared is equal to the product of the two lengths that it divides the base into.Now can you find out expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$

can you finish the problem……..

Now $x^2+2x-15=0$\(\Rightarrow (x-3)(x+5)=0\)\(\Rightarrow x=3,-5\) as length can not be negetive then \(x=3\).So area of \(\triangle ABC=60\)

Therefore the total Region \(ABCD\)= are of \(\triangle ABC\) + area of \(\triangle ACD\)=\(300+60\)=\(360\)

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Tetrahedron Problem | AMC-10A, 2011 | Problem 24

Try this beautiful problem from Geometry based on Tetrahedron.

Tetrahedron Problem – AMC-10A, 2011- Problem 24


Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?

  • \(4\sqrt 5+3\sqrt 2\)
  • \(4\sqrt 5+3\sqrt 2\)
  • \(\frac{1}{6}\)
  • \(7\sqrt 3+4\sqrt 2\)
  • \(\frac{1}{8}\)

Key Concepts


Geometry

Tetrahedron

Pythagoras

Check the Answer


Answer: \(\frac{1}{6}\)

AMC-10A (2011) Problem 24

Pre College Mathematics

Try with Hints


Tetrahedron Problem

We know that if we split a regular tetrahedron then it will split into eight tetrahedra that have lengths of $\frac{1}{2}$. The volume of a regular tetrahedron can be found using base area and height. Let us assume that the side length of the above tetrahedron is 1, its base area is $\frac{\sqrt{3}}{4}$, and its height can be obtaine using Pythagoras’ Theorem.Can you find out the height …..

can you finish the problem……..

Tetrahedron Problem

Therefore the height will be \(\sqrt{1^2-(\frac{\sqrt 3}{3})^2}\)=\(\frac{\sqrt2}{\sqrt 3}\) And the volume will be

\(\frac{1}{3} .\frac{\sqrt 3}{4}.\frac{\sqrt 2}{\sqrt 3\)=\(\frac{\sqrt 2}{12}\)

Now the actual side length of the Tetrahedron=\(\sqrt 2\).Therefore the actual volume is \(\frac{\sqrt 2}{12}.{\sqrt 2}^3=\frac{1}{3}\)

can you finish the problem……..

Tetrahedron Problem

There are eight small tetrahedron, the four tetrahedra on the corners of the large tetrahedra are not inside the other large tetrahedra. Thus, $\frac{4}{8}=\frac{1}{2}$ of the large tetrahedra will not be inside the other large tetrahedra.

Therefore  the volume of the region formed by the intersection of the tetrahedron is \(\frac{1}{3} \times \frac{1}{3}\)=\(\frac{1}{6}\)

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Problem on Equilateral Triangle | AMC-10A, 2010 | Problem 14

Try this beautiful Geometry Problem on Equilateral Triangle from AMC-10A, 2010.

Equilateral Triangle – AMC-10A, 2010- Problem 14


Triangle $ABC$ has $AB=2 \cdot AC$. Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$, respectively, such that $\angle BAE = \angle ACD$. Let $F$ be the intersection of segments $AE$ and $CD$, and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$?

  • \(60^{\circ}\)
  • \(70^{\circ}\)
  • \(90^{\circ}\)
  • \(75^{\circ}\)
  • \(1200^{\circ}\)

Key Concepts


Geometry

Triangle

Angle

Check the Answer


Answer: \(90^{\circ}\)

AMC-10A (2010) Problem 14

Pre College Mathematics

Try with Hints


Problem on equilateral triangle

We have to find out the \(\angle ACB\).Given that \(\angle CEF\) is a equilateral triangle and also given that $\angle BAE = \angle ACD$.so using the help of this two conditions ,we can find out all possible values of angles………

can you finish the problem……..

triangle figure

\(\angle BAE=\angle ACD=X\)

Let,

\(\angle BAE=\angle ACD=X\)

\(\angle BCD=\angle AEC=60^{\circ}\)

\(\angle EAC +\angle FCA+ \angle ECF+\angle AEC=\angle EAC +x+60^{\circ}+60^{\circ}=180^{\circ}\)

\(\angle EAC=60^{\circ}-x\)

\(\angle BAC =\angle EAC +\angle BAE =60^{\circ} -x+x=60^{\circ}\)

can you finish the problem……..

Since \(\frac{AC}{AB}=\frac{1}{2} \angle BCA\)=\(90^{\circ}\)

Therefore value of \(\angle BCA=90^{\circ}\)

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AMC 8 Math Olympiad USA Math Olympiad

Hexagon Problem | Geometry | AMC-10A, 2010 | Problem 19

Try this beautiful problem on area of Hexagon from Geometry.

Hexagon Problem – AMC-10A, 2010- Problem 19


Equiangular hexagon \(ABCDEF\) has side lengths \(AB=CD=EF=1\) and \(BC=DE=FA=r\). The area of \(\triangle ACE\) is \(70\%\) of the area of the hexagon. What is the sum of all possible values of \(r\)?

  • \(6\)
  • \(\frac{2}{\sqrt5} \)
  • \(8\)
  • \(9\)
  • \(9\)

Key Concepts


Geometry

Hexagon

Triangle

Check the Answer


Answer: \(6\)

AMC-10A (2010) Problem 19

Pre College Mathematics

Try with Hints


Hexagon Problem - figure

Given that area of \(\triangle ACE\) is \(70\)% of the area of the hexagon.so at first we have to find out the area of the \(\triangle ACE\).Clearly \(\triangle ACE\) is an equilateral triangle.Now from the cosines law we can say that \(AC^2=r^2+1^2-2r cos \frac{2\pi}{3}=r^2+r+1\).Therefore area of \(\triangle ACE=\frac{\sqrt 3}{4}(r^2+r+1)\).Can you find out area of the Hexagon \(ABCDEF\)?

Can you now finish the problem ……….

Area Of The Hexagon \(ABCDEF\) :

Hexagon in a triangle

If we extend $BC$, $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$, $BC$ and $DE$ meet at $Y$, and $DE$ and $FA$ meet at $Z$,Now we can find out area of the hexagon $ABCDEF$ which is formed by taking equilateral triangle $XYZ$ of side length $r+2$. So if we remove three equilateral triangles, $ABX$, $CDY$ and $EFZ$, of side length $1$.

Shaded hexagon figure

Therefore The area of $ABCDEF$ is \(\frac{\sqrt 3}{4} (r+2)^2 – \frac{3\sqrt 3}{4}=\frac{\sqrt 3}{4} (r^2+4r+1)\)

can you finish the problem……..

Given that area of \(\triangle ACE\) is \(70\)% of the area of the hexagon.Therefore

\(\frac{\sqrt 3}{4}(r^2+r+1)\)=\(\frac{7}{10}.\frac{\sqrt 3}{4}(r^2+4r+1)\)

\(\Rightarrow r^2-6r+1=0\).Now from Vieta’s Relation the sum of the possible value of \(r\) is \(6\)

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AMC 8 Math Olympiad USA Math Olympiad

Problem on Area of Trapezoid | AMC-10A, 2002 | Problem 25

Try this beautiful problem on area of trapezoid from Geometry.

Problem on Area of Trapezoid – AMC-10A, 2002- Problem 25


In trapezoid \(ABCD\) with bases \(AB\) and \(CD\), we have \(AB = 52\), \(BC = 12\), \(CD = 39\), and \(DA = 5\). The area of \(ABCD\) is

  • \(182\)
  • \(195\)
  • \(210\)
  • \(234\)
  • \(260\)

Key Concepts


Geometry

Trapezoid

Triangle

Check the Answer


Answer: \(210\)

AMC-10A (2002) Problem 25

Pre College Mathematics

Try with Hints


Area of Trapezoid - Problem

Given that \(ABCD\) is a Trapezium with bases \(AB\) and \(CD\), we have \(AB = 52\), \(BC = 12\), \(CD = 39\), and \(DA = 5\).we have to find out the area of the Trapezium.Normally the area of the trapezium is \(\frac{1}{2} (AD +BC) \times \)(height between CD & AB).but we don’t know the height.So another way if we extend \(AD\) & \(BC\) ,they will meet a point \(E\).Now clearly Area of \(ABCD\)=Area of \(\triangle ABE\) – Area of \(\triangle EDC\).Can you find out the area of \(\triangle EAB\) & Area of \(\triangle EDC\)?

Can you now finish the problem ……….

Problem on Area of Trapezoid

Now \(AB||DC\) , Therefore \(\triangle EDC \sim \triangle EAB\)

\(\Rightarrow \frac{ED}{EA}=\frac{EC}{EB}=\frac{DC}{AB}\)

\(\Rightarrow \frac{ED}{ED+DA}=\frac{EC}{EC+BC}=\frac{DC}{AB}\)

\(\Rightarrow \frac{ED}{ED+5}=\frac{EC}{EC+12}=\frac{39}{52}\)

Now , \(\frac{ED}{ED+5}=\frac{39}{52}\)

\(\Rightarrow ED=15\)

And \( \frac{EC}{EC+12}=\frac{39}{52}\)

\(\Rightarrow CE=36\)

Therefore \(BE\)=\(12+36=48\) and \(AE=20\)

Notice that in the \(\triangle EDC\), \({ED}^2 +{EC}^2=(36)^2+(15)^2=(39)^2=(DC)^2\) \(\Rightarrow \triangle EDC\) is a Right-angle Triangle

Therefore Area of \(\triangle EDC=\frac{1}{2} \times 36 \times 15=270\)

Similarly In the \(\triangle EAB\), \({EA}^2 +{EB}^2=(48)^2+(20)^2=(52)^2=(AB)^2\) \(\Rightarrow \triangle EAB\) is a Right-angle Triangle

Therefore Area of \(\triangle EAB=\frac{1}{2} \times 48 \times 20=480\)

Now can you find out the area of \(ABCD\)?

can you finish the problem……..

Shaded Area of Trapezoid

Therefore Area of \(ABCD\)=Area of \(\triangle ABE\) – Area of \(\triangle EDC\)=\(480-270=210\)

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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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