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Geometry Math Olympiad USA Math Olympiad

Centroid Problem: Ratio of the areas of two Triangles

Try this beautiful problem from Geometry based on Centroid.

Centroid Problem: Ratio of the areas of two Triangles


\(\triangle ABC\) has centroid \(G\).\(\triangle ABG\),\(triangle BCG\), and \(\triangle CAG\) have centroids \(G_1\), \(G_2\), \(G_3\) respectively. The value of \(\frac{[G1G2G3]}{[ABC]}\) can BE represented by \(\frac{p}{q}\) for positive integers \(p\) and \(q\).

Find \(p+q\) where\([ABCD]\) denotes the area of ABCD.

  • $14$
  • $ 10$
  • $7$

Key Concepts


Geometry

Triangle

centroid

Check the Answer


Answer: \(10\)

Question Papers

Pre College Mathematics

Try with Hints


Centroid Problem

\(\triangle ABC\) has centroid \(G\).\(\triangle ABG\),\(triangle BCG\), and \(\triangle CAG\) have centroids \(G1\),\(G2\),\(G3\) respectively.we have to find out value of \(\frac{[G1G2G3]}{[ABC]}\) i.e area of \(\frac{[G1G2G3]}{[ABC]}\)

Let D, E, F be the midpoints of BC, CA, AB respectively.
Area of $\frac{[DEF]}{[ABC]}$=$\frac{1}{4}$

we know that any median is divided at the centroid $2:1$. Now can you find out \(GG_1,GG_2,GG_3\) ?

Can you now finish the problem ……….

Centroid Problem

we know that any median is divided at the centroid $2:1$
Now  $G_1$ is the centroid of $\triangle ABG$, then$GG_1=2G_1F$
Similarly,$GG_2 = 2G_2D$ and$GG_3 = 2G_3E$
Thus, From  homothetic transformation  $\triangle G_1G_2G_3$ maps to $\triangle FDE$ by a homothety of ratio$\frac{2}{3}$
Therefore,$\frac{[G_1G_2G_3]}{[DEF]}$ = $(\frac{2}{3})^2$=$\frac{4}{9}$

can you finish the problem……..

Therefore we say that $\frac{[G_1G_2G_3]}{[ABC]}$ = $\frac{[G_1G_2G_3]}{[DEF]}\cdot \frac{[DEF]}{[ABC]} $= $\frac{4}{9}\cdot \frac{1}{4}$=$\frac{1}{9}$=$\frac{p}{q}$

So $p+q$=$9+1$=\(10\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Area of Trapezium | AMC-10A, 2018 | Problem 24

Try this beautiful problem from Geometry based on the Area of the Trapezium

Area of the Trapezium – AMC-10A, 2018- Problem 24


Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

  • $79$
  • $ 75$
  • $82$

Key Concepts


Geometry

Triangle

Trapezium

Check the Answer


Answer: $75$

AMC-10A (2018) Problem 24

Pre College Mathematics

Try with Hints


Area of Trapezium - Problem

We have to find out the area of BGFD.Given that AG is the angle bisector of \(\angle BAC\) ,\(D\) and \(E\) are the mid points of \(AB\) and \(AC\). so we may say that \(DE ||BC\) by mid point theorm…

So clearly BGFD is a Trapezium.now area of the trapezium=\(\frac{1}{2} (BG+DF) \times height betwween DF and BG\)

can you find out the value of \(BG,DF \) and height between them….?

Can you now finish the problem ……….

Area of Trapezium - figure

Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$.

Therefore area of \(\triangle ABC\)=\(\frac{ah}{2}\)=\(120\)………………..(1)

From the angle bisector theorem, we have that\(\frac{50}{x} = \frac{10}{y}\) i.e \(\frac{x}{y}=5\)

Let \(BC\)=\(a\) then \(BG\)=\(\frac{5a}{6}\) and \(DF\)=\(\frac{1}{2 } \times BG\) i.e \(\frac{5a}{12}\)

now can you find out the area of Trapezium ?

can you finish the problem……..

Area of shaded Trapezium

Therefore area of the Trapezium=\(\frac{1}{2} (BG+DF) \times FG\)=\(\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}\)=\(\frac{ah}{2} \times \frac{15}{24}\)=\(120 \times \frac{15}{24}\)=\(75\) \((from ……..(1))\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Area of the Octagon | AMC-10A, 2005 | Problem 20

Try this beautiful problem from Geometry based on the Area of the Octagon from AMC 10A, 2005, Problem 2005

Area of the octagon – AMC-10A, 2005- Problem 20


An equiangular octagon has four sides of length 1 and four sides of length \(\frac{\sqrt{2}}{2}\), arranged so that no two consecutive sides have the same length. What is the area of the octagon?

  • \(\frac{4+5\sqrt 2}{2}\)
  • \(\frac{7}{2}\)
  • \(7\)

Key Concepts


Geometry

Triangle

Octagon

Check the Answer


Answer: \(\frac{7}{2}\)

AMC-10A (2005) Problem 20

Pre College Mathematics

Try with Hints


Octagon figure

We have to find out the equiangular octagon whose four sides of length 1 and four sides of length \(\frac{\sqrt{2}}{2}\),

we join \(AD\),\(HE\),\(BG\) and \(CF\).We assume that side lengths of \(AB=CD=EF=GH=1\) and side lengths of \(AH=BC=DE=GF=\frac{\sqrt{2}}{2}\)( As no two consecutive sides have the same length). Now

Can you now finish the problem ……….

finding the Area of Octagon


There are 5 squares with side lengths \(\frac{\sqrt{2}}{2}\) and 4 Triangles of side lengths \(1\)

Now area of \(5\) squares=\( 5 \times (\frac{\sqrt{2}}{2})^2\)=\(\frac{5}{2}\) and area of each Triangle is half of the area of a square.so the area of \(4\) Triangles=\(4 \times \frac{1}{2} \times \frac{1}{2}\)=\(1\)

can you finish the problem……..

Therefore the Total area of the required octagon=Total area of Five squares + Total areas of Four Triangles=\(\frac{5}{2} +1\)=\(\frac{7}{2}\)

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AMC 8 Math Olympiad USA Math Olympiad

Area of the Trapezium | AMC-10A, 2018 | Problem 24

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

Area of the Trapezium – AMC-10A, 2018- Problem 24


Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

  • $79$
  • $ 75$
  • $82$

Key Concepts


Geometry

Triangle

Trapezium

Check the Answer


Answer: $75$

AMC-10A (2018) Problem 24

Pre College Mathematics

Try with Hints


Area of the Trapezium

We have to find out the area of BGFD.Given that AG is the angle bisector of \(\angle BAC\) ,\(D\) and \(E\) are the mid points of \(AB\) and \(AC\). so we may say that \(DE ||BC\) by mid point theorm…

So clearly BGFD is a Trapezium.now area of the trapezium=\(\frac{1}{2} (BG+DF) \times height betwween DF and BG\)

can you find out the value of \(BG,DF \) and height between them….?

Can you now finish the problem ……….

Area of the Trapezium- Problem

Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$.

Therefore area of \(\triangle ABC\)=\(\frac{ah}{2}\)=\(120\)………………..(1)

From the angle bisector theorem, we have that\(\frac{50}{x} = \frac{10}{y}\) i.e \(\frac{x}{y}=5\)

Let \(BC\)=\(a\) then \(BG\)=\(\frac{5a}{6}\) and \(DF\)=\(\frac{1}{2 } \times BG\) i.e \(\frac{5a}{12}\)

now can you find out the area of Trapezium and area of Triangle?

can you finish the problem……..

Area of the shaded portion

Therefore area of the Trapezium=\(\frac{1}{2} (BG+DF) \times FG\)=\(\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}\)=\(\frac{ah}{2} \times \frac{15}{24}\)=\(120 \times \frac{15}{24}\)=\(75\) \((from ……..(1))\)

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