4

We have an isosceles triangle and we divide it by two sections going out of one of three corners, hence we get three new triangles. Is it possible to make (puzzle) an isosceles triangle out of every two of three triangles that we got after dividing the first one? (In case you would ask me what I tried, I begged my teacher for an answer. Yes, it is possible, but the proof is still needed) Thank you in advance!

Severus156
  • 351

2 Answers2

1

I'm not sure if I get the Question right, but it should not, in general, be possible (I don't know how to make a picture here, so I will try to explain it as best as possible).

Take an icosceles triangle, call it $\Delta(A,B,C)$ where $A,B \quad \text{and} \quad C$ are the names of the corners. Now, let's start with Point C, and draw 2 lines from there which will intersect $\overline{AB}$ at some Points $C_1$ and $C_2$.

Take the corresponding triangles $\Delta(A,C_1,C)$, $\Delta(C_1,C_2,C)$ and $\Delta(C_2,B,C)$. Now, If I understand you correctly, you want to know if it is possible to ''glue'' any two of these two triangles together and get an icosceles triangle.

In general, this should not be possible, take for example $\Delta(A,C_1,C)$ and $\Delta(C_1,C_2,C)$, we know that the lenght $L(\overline{CC_1})$ of $\Delta(A,C_1,C)$ equals $L(\overline{CC_1})$ of $\Delta(C_1,C_2,C)$ but nothing about the remaining side-lenghts. In particular, you can (try it at home!) divide it in such a way, that $L(\overline{AC_1}) \neq L(\overline{C_1C_2})$ and $L(\overline{CA}) \neq L(\overline{CC_2})$. Thus, the only possible way to glue the two of these together would be to glue at the $\overline{CC_1}$ line! But this results in the triangle $\Delta(AC_2C)$ hence, if you split $\Delta(A,B,C)$ in such a way that $L(\overline{AC_2}) \neq L(\overline{AC})$ there is no possible way to acieve your goal (the latter will always be the case since your triangle was icosceles so any nontrivial choice of sections will give this result).

Creo
  • 419
  • 3
  • 6
  • I think he means that you can choose to how to cut up your isosceles triangle i.e., you can choose your lines they are not arbitrary. In that case, it looks like it would be possible. – user625 Jun 17 '18 at 17:53
  • Exactly, you can choose how to cut up a given isosceles triangle. – Severus156 Jun 17 '18 at 23:39
  • See edit :) (Didn't have enough space here) – Creo Jun 18 '18 at 07:40
  • Hmm, I did one case below. It does look like it would work. Let triangle ABC have two equal sides AB and AC of length $a$ and base BC of length $b$.

    Case 1: $b \gt a$

    In this case mark two points on BC at D and E so that BD = a and CE = a. Note that BC < 2a, by the triangle inequality so $BD > b/2$ (think why this is essential for the construction to work). Now, cut the triangle into 3 parts: ABE, ACD and ADE. Join ABE, ADE along AE; ACD and ADE along AD and ABE, ACD along BE = DC. These three are isosceles triangles (why?)

     – user625 Jun 18 '18 at 19:25
  • Ah, yeah my bad, I thought of a triangle where all lenghts are equal! (My english wasn't as on point as I thought... I will reedit my comment) – Creo Jun 18 '18 at 22:26
  • @user625: Wouldn't joining your $\triangle ABE$ and $\triangle ACD$ along $\overline{BE}\cong\overline{DC}$ create a quadrilateral (either a parallelogram or a concave kite)? – Blue Jun 19 '18 at 06:10
  • 1
    @Blue you are right. It doesnt work for the reason you gave. – user625 Jun 19 '18 at 11:19
1

It is possible (at least for some isosceles triangles).

One isosceles to two isosceles triangles

Take an isoceles triangle $ABC$ with $AB = AC$ and $36^\circ < \angle BAC < 45^\circ$.

  • Let the perpendicular bisector of $AC$ intersect $AB$ at $D$.
  • Draw a circle at $C$ through $D$, it will intersect $AB$ at $D$ and $E$.
  • Cut $\triangle ABC$ along $CD$ and $CE$, we get two new isosceles $\triangle ADC$ and $\triangle CDE$
    (by construction, $AD = CD = CE$ ).
achille hui
  • 122,701
  • I think this nice construction may work for all isosceles triangles if you are allowed to subtract triangles as well as add them. (Not what the OP meant, of course.) – Ethan Bolker Jun 19 '18 at 00:25
  • Isn't the goal to create three sub-triangles that we can join pairwise to form three isosceles triangles? – Blue Jun 19 '18 at 06:12