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All sides of a triangle T1 are shorter than the shortest side of a triangle T2. Is it always possible to put triangle T1 completely inside triangle T2?

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The answer is

no

because, for example, if T2 has side-lengths

50, 50, 99 (so T2 is isosceles and very flat, almost no area at all, one angle near 180 degrees and the other two near 0),

then a triangle T1 such as

the equilateral triangle of side-lengths 20, 20, 20 will not fit inside T2, because the height of T2 is so low.

More generally, we can consider T2 with side-lengths

$n$, $n$, $2n-1$ for any large $n$ (when $n$ is very large, the angles of this rectangle tend towards $0,0,180$ degrees in the infinite limit)

and T1 with side-lengths

$n-1$, $n-1$, $n-1$, which clearly won't fit inside the almost-completely-flat triangle when $n$ is large.

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The answer is

no.

Let T2 be a triangle with sides 20, 11 and 11. We can calculate its area with Heron's formula; $s = \frac{20 + 11 + 11}{2} = 21$, so the area is $\sqrt{21 \cdot 1 \cdot 10 \cdot 10} \approx 45.8$. The area is also equal to half the height $h$ times the base ($20$); that means the height is $4.58$. An equilateral triangle T1 with all sides length 10 clearly won't fit in this T2.

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  • $\begingroup$ Are you sure this is true? It seems to me that, putting the shortest side at the bottom, the width of T2 would be close to 1.1 for quite a way above its shortest side, and so T1 would have room to fit in there. $\endgroup$ – Rand al'Thor Jan 20 at 20:40
  • $\begingroup$ Sorry, right formula, wrong idea. $\endgroup$ – Glorfindel Jan 20 at 20:54
  • $\begingroup$ Now the idea is the same as my answer. You've included a rigorous proof that T1 won't fit in T2, but I've taken even bigger numbers to make it "clearer". In fact, by a limiting process it can be completely clear ... brb editing. $\endgroup$ – Rand al'Thor Jan 20 at 21:03
  • $\begingroup$ My browser missed that you posted an answer as well :) $\endgroup$ – Glorfindel Jan 20 at 21:04

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