# What's the perimeter of this poorly specified triangle? [duplicate]

Generalizing a puzzle from Mind Your Decisions, here's something that I found to be rather neat.

Suppose that AB$$=c$$, AC$$=b$$, and BC$$=a$$. What's the perimeter of $$\triangle$$CDE?

Clue: The coveted tick will go to the most attractive, visual proofs/arguments that don't use any additional variables! Judges decision etc.

• Do you mean $BC=a$ ? – hexomino Feb 14 '19 at 15:57

It is probably not something OP expects (with a lot of additional variables) but I would like to publish my idea anyway: The clue is to "build" |AD| and |BE| segments from fragments of |AB| and |DE|

(we can also use a known property of quadrilateral circumscribing circle which says that |AD| + |BE| = |AB| + |DE|)

We can derive that:

a + b = (|AD| + |DC|) + (|CE| + |EB|) = c + |DE| + |DC| + |EC|

so by subtracting c from both sides:

|DE| + |DC| + |EC| = a + b - c

Since $$AB=BC=c$$, and $$AC=b$$ we know this is an isosceles triangle (which makes the diagram a poor representation of the question). This means that the point where the inscribed circle touches $$AC$$ (the side not equal to the others) is the midway mark. Lets call this point $$M$$.

Since no guidance was given on $$D$$ and $$E$$, one can move $$D$$ all the way towards $$M$$. To maintain the tangent, $$E$$ then nears $$C$$. As $$D$$ approaches $$M$$, and $$E$$ approaches $$C$$, the triangle approaches an isosceles triangle whose two long sides are both $$DC$$ and whose short side is $$0$$. Thus, the perimeter of $$CDE$$ is $$CD+CD=2CD$$. Since $$AC=c$$ and $$CD=CM$$ is half of $$AC$$, we know $$CD=\frac{c}{2}$$.

Thus, The perimeter of $$CDE=2CD=2\frac{c}{2}=c$$.

• This was probably correct, but the OP had actually made a typo. AB != BC, and hence, it is not an isosceles triangle. – Aryaman Feb 16 '19 at 7:00