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So I've encountered a question where it requires me to find the maximum area of a small right angled triangle inside a bigger one. The question stated the dimensions of the big triangle and approved the parallelism of 2 lines in the form below: enter image description here

Everything in black is given by the question, and otherwise (red) is assumed by me.

I've assumed the line $\overleftrightarrow{eb}$ equal to $x$ and $\overleftrightarrow{db}$ equal to y. Then I proved that $\Delta$ $deb$ is similar to $\Delta$ $abc$ by sharing the same right angle and having the angle $<deb$ corresponding to angle $<acb$, thus:

$$\frac{y}{6} = \frac{x}{8}$$, then $$ y = \frac{3x}{4} $$.

$$\overleftrightarrow{de}$$ would be equal to $$\frac{5x}{4} $$

Now I had to find the height $$\overleftrightarrow{fe}$$ in terms of $$x$$

Since $$\overleftrightarrow{de}$$ is parallel to $$\overleftrightarrow{ac}$$ , angles $$<edf$$ and $$<afd$$ are alternate angles, thus they are equal to each other, and since $$<afd$$ and $$<acb$$ are corresponding angles, $$<edf$$ is equal to $$<acb$$, and both $$\Delta abc, \Delta fed$$ have right angles, then we can infer that $$<efd$$ is equal to $$<cab$$, so both triangles are similar.

Thus: $$\frac{n}{6} = \frac{\frac{5x}{4}}{8}$$ and $$n = \frac{15x}{16} $$.

Now we can find the area of the smaller triangle as a function of $x$.

$$f(x) = 0.5 × \frac{15x}{16} × \frac{5x}{4}$$

But, if I were to take the derivative of that function to find a maximum value, I would end up with a minimum value at $x = 0$, which is utterly irrational. What mistake have I done here?

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  • $\begingroup$ Pardon, I missed it, it's fde $\endgroup$ Jul 22, 2018 at 12:59
  • $\begingroup$ I can see the triangle fde in your figure; but you have not clearly defined the set of admissible small triangles. As it stands the outer triangle does qualify, hence is trivially the largest. $\endgroup$ Jul 22, 2018 at 13:05
  • $\begingroup$ Nope, the question stated the maximum possible area of fde only. $\endgroup$ Jul 22, 2018 at 13:44

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Hint:let $$ED=m$$ then we get: $$A=\frac{1}{2}mn$$ where $$n=\frac{3}{5}(8-x)$$ and $$m=\sqrt{x^2+\left(\frac{3}{4}x\right)^2}$$ We get then $$A=\frac{3}{8}(8x-x^2)$$ And the maximum we get for $$x=4$$

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  • $\begingroup$ I don't get it how did y become (3/5x) ? I mean, is my definition of y (y = 3x/4) incorrect? Also how did you get n? Can you elaborate please? Maybe I have some geometrical mistake $\endgroup$ Jul 22, 2018 at 11:31
  • $\begingroup$ Yes it is $$\frac{y}{x}=\frac{3}{4}$$ just a typo! $\endgroup$ Jul 22, 2018 at 11:46
  • $\begingroup$ $$\sin(\alpha)=\frac{n}{8-x}=\frac{6}{\sqrt{6^2+8^2}}$$ and $$\alpha$$ denotes the angle at $$C$$ $\endgroup$ Jul 22, 2018 at 11:50
  • $\begingroup$ it is better now? $\endgroup$ Jul 22, 2018 at 11:52
  • $\begingroup$ Note that the angle at $f$ is $\frac{\pi}{2}$ $\endgroup$ Jul 22, 2018 at 11:53
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$ \overline{de} = \frac{5}{4}x $ as you said,

since triangle $ cef$ and triangle $cab$ are similar triangles,

$ \overline{ce} = 8-x $

$ 8-x:n=5:3 $

$5n = 24-3x$

$n = \frac{24-3x}{5}$

$A = \frac{24-3x}{5}\times\frac{5}{4}x\times\frac{1}{2} = \frac{24x-3x^2}{4}\times\frac{1}{2} =3x-\frac{3}{8}x^2$

$\frac{\text{d}A}{\text{d}x} = 3 - \frac{3}{4}x = 0$

$x = 4$

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  • $\begingroup$ Yeah, I see that you came with another method which works too, by using the similarity of two other triangles. But I have a question; why does my method give different answer? As you see, I used the theorem of triangles similarity twice, first time everything went ok, but the second time when I used it between triangles abc and fde, in order to fetch out "n", it gave me unsatisfying results. Utterly different. Why? $\endgroup$ Jul 22, 2018 at 12:20
  • $\begingroup$ @StephenAlexander because $\overline{df}$ and $\overline{bc}$ are not parallel $\endgroup$
    – Pizzaroot
    Jul 22, 2018 at 12:27
  • $\begingroup$ Hmm so you are implying that adf and abc aren't similar triangles? Makes sense now. So let's say that if df was connecting the midpoints of ab and ac then it would be parallel with the third line right? Otherwise any line that comes above or below isn't parallel to the third line? $\endgroup$ Jul 22, 2018 at 12:37
  • $\begingroup$ yeah but it wouldn't be the midpoints of ab and ac since the diagram only showed that ac and de are parallel and the angle def is right angle. df and bc could be parallel, but it wasn't mentioned in the question $\endgroup$
    – Pizzaroot
    Jul 22, 2018 at 12:46
  • $\begingroup$ True. It doesn't even connect the midpoints too. Appreciated $\endgroup$ Jul 22, 2018 at 12:53
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enter image description here

Let $|BC|=a=8$cm, $|AB|=c=6$cm, then $|AC|=b=10$cm.

\begin{align} S_{\triangle FED}(x)&=\tfrac12|DE|\cdot|FE| \\ &=\tfrac12\cdot\frac{x}{\cos\gamma}\cdot(a-x)\cos\alpha \\ &= \tfrac12\cdot\frac{x}{\tfrac{a}{b}}\cdot(a-x)\cdot\tfrac{c}{b} \\ S_{\triangle FED}(x)&=\frac{c}{2a}\,(ax-x^2) . \end{align}

\begin{align} S'_{\triangle FED}(x)&= \frac{c}{2a}\,(a-2x) ,\\ S''_{\triangle FED}(x)&= -\frac{c}{a} <0\quad \forall x , \end{align}

hence $x=\frac{a}2$ provides the maximum

\begin{align} S_{\triangle FED}(x)_{\max}&= S_{\triangle FED}(\tfrac{a}2) =\frac{ac}8 =6\,\mathrm{cm}^2 . \end{align}

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  • $\begingroup$ Quick and neat method. $\endgroup$ Jul 22, 2018 at 17:27

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