Suppose you have the following triangle $ABC$:enter image description here

with the following properties: $|AB|=4\cdot |AA'|$, $|AC|=4\cdot |CC'|$, $|BC|=4\cdot |BB'|$. I have to find the ratio of the total area of the triangle and the red area. I tried alot of algebraic manipulations with the lengths of the sides, but couldnt solve it that way. I really need hints for this one. Thanks!

  • $\begingroup$ If it helps, I found empirically that the ratio is very nearly 13:4. $\endgroup$ – David H Aug 27 '13 at 16:04
  • $\begingroup$ I don't know if this counts as a 'hint' or not, but the fact that the problem doesn't specify the lengths of $AB$, $AC$ or $BC$ suggests that it's independent of the size of the original triangle, and this suggests taking e.g. $ABC$ as a 45-45-90 right triangle... $\endgroup$ – Steven Stadnicki Aug 28 '13 at 23:40
  • $\begingroup$ The general version of this question appears here: math.stackexchange.com/q/261819/409 along with my answer: math.stackexchange.com/a/261870/409 . $\endgroup$ – Blue Aug 29 '13 at 5:49

EDIT: Final solution.

First off, notice that $\triangle ABB',\triangle BCC', \triangle CAA'$ all have area equal to $\frac{1}{4}$ of the area of $\triangle ABC$. To see this, pick any side as the base when measuring $\triangle ABC$, say side $BC$. Then $BB'$ is a triangle with a common height and one fourth base, so its area is one fourth. This argument obviously holds for all three sides.

What might be more surprising is that all three of the small triangles at the corners have the same area. They all equal $\frac{1}{13}\times \frac{1}{4}$ of the area of $\triangle ABC$, or rather, area equal to one thirteenth of the three aforementioned triangles. To make this explicit, look at $\triangle ABB'$ and $\triangle AA'P$, where $P$ is the intersection of $A'C$ and $AB'$. We have the following relationships: $$AB = 4AA' \text{, and }\angle{BAB'} = \angle{A'AP} = \theta$$ Next we have the following equations for the triangles' areas: $$\triangle AA'P = \frac{1}{2}AP\times AA'\sin\theta \text{, and } \triangle ABB' = \frac{1}{2}AB\times AB'\sin\theta $$ Therefore, the ratio of their areas is $\dfrac{\triangle AA'P}{\triangle ABB'} = \dfrac{AP}{4AB'}$. Now as soon as I can prove that this last quantity equals $\frac{1}{13}$, the proof will be complete.

EDIT2: Got it.

My solution requires Menelaus' Theorem. I will not prove it here.

By Menelaus' Theorem, we have that $$\begin{align*} BC \times B'P \times AA' &= B'C \times AP \times BA' \\ BC \times B'P \times \frac{1}{4}AB &= \frac{3}{4}BC \times AP \times \frac{3}{4}AB \\ \dfrac{B'P}{AP} &= \dfrac{9}{4}\\ \dfrac{AB' - AP}{AP} &= \dfrac{9}{4}\\ \dfrac{AB'}{AP} &= \dfrac{13}{4} \implies \dfrac{AP}{AB'} = \dfrac{4}{13}\\ \end{align*}$$

Thus, as desired $\dfrac{AP}{4AB'} = \dfrac{1}{13}$. Thus, the area of the central triangle is $$\triangle ABC(1 - 3\times \frac{1}{4} + 3 \times \frac{1}{13}\times \frac{1}{4}) = \triangle ABC \times \frac{4}{13}$$

This is my first solution using trigonometry and the particular case of an equiangular triangle. According to Blue, due to affine equivalence of all triangles, this proof suffices for all cases.

Let $AB = BC = AC = 4k$, and hence $AA'=BB'=CC'=k$ and $\angle{ABC} = \angle{BCA} = \angle{BAC} = 60$.

enter image description here

I claim the central triangle is equiangular. Examine one of the three small triangles at the corners of $\triangle ABC$. For convenience, I'll focus on the top one. Call it $AA'P$, where $P$ is the intersection of $CA'$ and $AB'$. Note the following $$ \angle{A'AP} = \angle{A'AB}\text{, and } \\ \angle{AA'P} = \angle{A'AC} = \angle{BB'A}$$ Furthermore, we know that $\angle{A'AB} + \angle{BB'A} = 120$, since the sum of the angles of any triangle equals 180. By this same principle, $\angle{A'PA} = 60$. This is true for all the small triangles, and by the vertical angles theorem, the central triangle is equiangular.

Notice that triangle $PA'A$ is similar to triangle $BB'A$. Thus we can find the lengths of $PA$ and $PA'$. But first we must compute the length of $B'A$. Via the law of cosines, $B'A = \sqrt{(4k)^2 +k^2 - 2(k)(4k)\cos(60)} = k\sqrt{13}$. Now using properties of similar triangles, we have the following equality

$$\dfrac{PA}{AB}=\dfrac{A'A}{B'A} \implies PA=\dfrac{AB\times A'A}{B'A} = \dfrac{4k}{\sqrt{13}}$$

By a similar argument, we can calculate the length $PA' = > \dfrac{k}{\sqrt{13}}$. So, the side lengths of the central triangle all equal $k\sqrt{13} - \dfrac{5k}{\sqrt{13}} = \dfrac{8k}{\sqrt{13}}$. Finally, the ratio of the areas of two equilateral triangles equals the ratio of the squares of their sides, hence

$$\dfrac{\triangle Outer}{\triangle Inner} = \dfrac{16k^2}{\frac{64k^2}{13}} = \boxed{\dfrac{13}{4}}$$

  • $\begingroup$ thanks for this answer, but a proof that this holds for every triangle is still needed. $\endgroup$ – Badshah Aug 28 '13 at 20:20
  • $\begingroup$ I agree. I haven't spent much more time thinking about it, but I wonder if a calculus based argument to be applied to this current configuration, by parameterizing the ratio of two triangles and demonstrating a zero derivative with respect to changes in the parameters. I can't help feeling there is a more geometric approach, though. $\endgroup$ – A.E Aug 28 '13 at 22:21
  • $\begingroup$ I've edited my post with updates on a more general solution, but I am stuck at the crux of my argument. $\endgroup$ – A.E Aug 28 '13 at 23:30
  • $\begingroup$ Ratios of parallel segments, and ratios of areas, are affinely independent. Since all triangles are affinely equivalent, solving the problem for one (non-degenerate) instance proves it for all of them. $\endgroup$ – Blue Aug 29 '13 at 5:40
  • 1
    $\begingroup$ Thank you very much, @Blue! $\endgroup$ – A.E Aug 29 '13 at 17:40

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