# Area of a triangle

The following problem in elementary geometry was proposed to me. As a mathematical analyst, I confess that I can't solve it. And I have no idea of what I could do. Here it is: pick a triangle, and draw the three mediana (i.e. the segments that join a vertex with the midpoint of the opposite side). Use the three segments to construct a second triangle, and prove that the area of this triangle is $3/4$ times the area of the original triangle.

Any help is welcome.

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What do you mean by "use the three segments "? Shouldn't it be "use the three new vertices" ? – DonAntonio Feb 3 '14 at 14:33
No: you draw the segment that joins a vertex with the opposite midpoint. You get three such segments, and you use them to build a triangle. – Siminore Feb 3 '14 at 14:36
Oh, I see...thanks. How can we be sure such a (n euclidean) triangle is constructible? – DonAntonio Feb 3 '14 at 14:38
Well, that is part of the problem. But I suspect it is not the hardest part. – Siminore Feb 3 '14 at 14:40
It should be $1/4$ instead of $3/4$. – Felix Marin Feb 3 '14 at 18:01

As suggested we can define any triangle with two vectors: $\mathbf{a}=\begin{pmatrix}{a_1 \\a_2} \end{pmatrix}$ and $\mathbf{b} = \mathbf{e}_1$, such that $\mathbf{a}$ and $\mathbf{b}$ are not colinear and where $\mathbf{b}$ has been chosen for simplicity.

Then a linear map, $A$, can be constructed to send these vectors to there corresponding line segments. If we get that the determinant of $A$ is such that $\det{A}=\frac{3}{4}$ the result will be proved.

After drawing a picture of a triangle defined in such a way it is clear that we want the map $A$ such that: $$\mathbf{a} \mapsto \mathbf{b} - \frac{1}{2}\mathbf{a} ,$$ $$\mathbf{b} \mapsto \mathbf{a} - \frac{1}{2}\mathbf{b} .$$

This completely determines $A$ and after solving some equations we get that: $$A=\begin{pmatrix}1-\frac{a_1}{2} & -\frac{1+a_1^2}{2a_2}\\-\frac{a_2}{2} & 1+\frac{a_1}{2} \end{pmatrix}.$$

Upon computing $\det(A)$ we get that it is $\frac{3}{4}$ proving what is required.

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I believe there is a nice visual explanation for this.

Then construct congruent triangles to form a hexagon, and connect every other vertex on its perimeter.

Let $A=$ (area of the original triangle) and $B=$ (area of the triangle of medians). The shaded triangle has area $4B$, because its sides are twice as long as the medians, and it is also equal to half the area of the hexagon, which is $6A/2=3A$. $3A=4B$, so $B=\frac{3}{4}A$.

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great answer! up-voted. – Mick Feb 17 '14 at 18:49
This is just awesome :) – Sawarnik Sep 8 '14 at 11:02

Use vectors.It will be really helpful.Define sides of triangle ABC as A=0, B= b vector and C= c vector.

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In a triangle $ABC$ with medians intersecting at $O$, draw a line throgh $A$ parallel to the median through $B$ and a line through $B$ parallel to the median through $A$. Let $D$ be the intersection of the new lines. Then the parallelogram $AOBD$ has area $2/3$ of the area of the triangle $ABC$ but it can also be partitioned by the diagonal $OD$ into two triangles made out of intervals of lengths $2/3$ of the lengths of the medians themselves. Hence the area of the triangle formed by the medians is $(3/2)^2$ times $1/3$, or $3/4$ of the area of the triangle $ABC$.

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Instead of "parallel to the side $BC$ (resp. $AC$)", don't you mean "parallel to the median through $B$ (resp. $A$)"? – TonyK Feb 3 '14 at 16:16
@TonyK, Right, sorry. – Mikhail Katz Feb 3 '14 at 16:19

Let $AA', BB', CC'$ be the medians, and let $G$ be their intersection. Let $G'$ be the symmetric of $G$ with respect to $A'$. Then, since $BGCG'$ is a parallelogram, in the triangle $BGG'$ we have

$$BG=\frac{2}{3}BB' \,;\, BG'=\frac{2}{3}CC' \,;\, GG'=\frac{2}{3}AA' \,.$$

Thus, area of $BGG'$ is $\frac{4}{9}$ area of the triangle made by the medians.

Also, area $BGG'=\frac{1}{2}$ area $BGCG'$=area $BGC=\frac{1}{3}$ area $ABC$.

[the last equality follows either by repeating the argument by the other two sides, or observing that $ABC$ and $GBC$ have the same base, and their heights are $3:1$].

Thus

$$\mbox{Area} BGG'=\frac{4}{9} \mbox{Area of the median triangle}$$

$$\mbox{Area} BGG'=\frac{1}{3} \mbox{Area} ABC$$

Simplification Here is a much simpler version of this proof:

Note the the area of $BGC$ is $\frac{1}{3}$ area of $ABC$.

Now cut $BGC$ along the side $GA'$ and rotate $CGA'$ by $180^o$ degrees. Then you get a triangle with the sides $\frac{2}{3}$ the medians. Equate the areas.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Let's $\vec{a}$, $\vec{b}$ and $\vec{c}$ the vertexs of the original triangle. Its area ${\cal A}$ is given by the magnitude of: $$\half\,\vec{a}\times\vec{b} + \half\,\vec{b}\times\vec{c} + \half\,\vec{c}\times\vec{a}$$ Middle points of the triangle are: $$\half\,\pars{\vec{a} + \vec{b}}\,,\ \half\,\pars{\vec{b} + \vec{c}}\ \mbox{and}\ \half\,\pars{\vec{c} + \vec{a}}$$ The 'new' area is given by: \begin{align} &\half\bracks{\half\,\pars{\vec{a} + \vec{b}}\times\half\,\pars{\vec{b} + \vec{c}}} + \half\bracks{\half\,\pars{\vec{b} + \vec{c}}\times\half\,\pars{\vec{c} + \vec{a}}} + \half\bracks{\half\,\pars{\vec{c} + \vec{a}}\times\half\,\pars{\vec{a} + \vec{b}}} \\[3mm]&= {\vec{a}\times\vec{b} + \vec{a}\times\vec{c} + \vec{b}\times\vec{c} \over 8} + {\vec{b}\times\vec{c} + \vec{b}\times\vec{a} + \vec{c}\times\vec{a} \over 8} + {\vec{c}\times\vec{a} + \vec{c}\times\vec{b} + \vec{a}\times\vec{b} \over 8} \\[3mm]&={\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a} \over 8} \\[3mm]&=\color{#00f}{\Large{1 \over 4}}\,\pars{{\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a} \over 2}} \end{align}

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The new triangle is not the one whose vertices are the midpoints of the old triangle's sides. It shall be a triangle whose sides have the lengths of the three medians. – Daniel Fischer Feb 3 '14 at 18:42
@DanielFischer 0 k. I'll checked it. Thanks. – Felix Marin Feb 3 '14 at 18:50

"Here it is: pick a triangle":

Since we get to pick the triangle, I will pick a triangle T1 that has equal sides, therefore 3 equal angles = 60 degrees. Let's call the sides "A". The area of this triangle T1 is equal to [0.5 * A * A]. Each of the 3 "medianas" will have a magnitude equal to [A * Sin(60)], therefore a triangle T2 constructed from these 3 medianas will have an area equal to (0.5 * [A * Sin(60)] * [A * Sin(60)])

Dividing the area of Triangle T2 by the area of triangle T1:

Ratio of Areas = (0.5 * [A * Sin(60)] * [A * Sin(60)]) / [0.5 * A * A]

Ratio of Areas = Sin(60) * Sin(60)

Ratio of Areas = 0.75 = 3/4

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