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#1 2014-06-19 06:36:17

harrychess
Member
Registered: 2014-04-04
Posts: 33

triangles

Let ABC be any triangle. Equilateral triangles BCX, ACY, and BAZ are constructed such that none of these triangles overlaps triangle ABC.

a) Draw a triangle ABC and then sketch the remainder of the figure. It will help if  ABC is not isosceles (or equilateral).

b) Show that, regardless of choice of  ABC, we always have AX = BY = CZ.

Problem A I can manage(obviously), but I dont know how to do B. If I made ABC an equilateral triangle, it would be easy, so I dont want to do that. Could you explain how to prove if ABC was obtuse scalene?

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#2 2014-06-19 12:30:11

Isomorph
Member
Registered: 2014-06-19
Posts: 5

Re: triangles

You can prove this by congruent triangles.

Notice that in triangles ACX and YCB, angle ACX = angle YCB, because each of those angles is formed by two adjacent angles: one being angle ACB and the other an angle from an equilateral triangle (60 degrees).

i.e. angle ACX = angle YCB = (angle ACB + 60).

Next, see that AC = YC (two sides of an equilateral triangle), and CX = CB for the same reason.

Therefore triangles ACX and YCB are congruent triangles, because they each have two sides which are respectively the same length, subtending the same angle.

It follows that AX = BY since they are corresponding sides.

You can apply the same type of argument to show that AX = CZ.

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#3 2014-06-20 00:59:02

Isomorph
Member
Registered: 2014-06-19
Posts: 5

Re: triangles

The above proof works for all triangles ABC, but it needs a small clarification when angle ACB > 120 degrees.

When angle ACB = 120 degrees, angles ACX and YCB become 180 degrees each, and the triangles ACX and YCB become trivial - they "collapse" into straight line segments of equal length. The proof still holds, and AX = YB (they are the line segments).

When angle ACB > 120 degrees, triangles ACX and YCB both now lie entirely outside  triangle ABC.
It is still true that angle ACX = angle YCB = (angle ACB + 60) but now we are talking about the EXTERNAL angles ACX and YCB (measuring them outside their triangles, the "long way round").

Therefore angle ACX = angle YCB (whether measured internally or externally) and the proof holds as before, with triangles ACX and YCB congruent.

Apologies for the lack of a diagram; I've only just joined and haven't learnt how to add diagrams yet. It should be easy to follow the above if you sketch it out.

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#4 2014-06-20 01:16:29

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,049

Re: triangles

Hi Isomorph

Welcome to the forum! smile


“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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#5 2014-06-21 12:04:12

Isomorph
Member
Registered: 2014-06-19
Posts: 5

Re: triangles

Hi anonimnystefy,

Thanks! Glad to be here; this looks like a nice place. smile

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