# If a triangle's area equals its perimeter, what special condition must its angles satisfy?

If we have a triangle where the Perimeter >0 and the Area >0 , and Area=Perimeter, what special condition must the angles of this triangle satisfy for this to happen?

I've done a bit of research and have found information on Heron's forumla and on integral triangles but I'm at a loss at how to figure this problem out.

Through research I have found that if the lengths of the sides of the triangle are either {6,8,10),(5,12,13),(9,10,17),(7,15,20), or(6,25,29) we get a triangle that fits the requirements (im not sure if this information is relevant) but the question is about the "special condition" that the angles must meet to create such a triangle not the sides.

I'm looking for both an answer and an explanation, any help is greatly appreciated. Please let me know if I left anything out (that's all the information I was given) or if something is unclear.

• Do you only care about triangles with integral sides? Sep 26, 2013 at 23:15
• No, the question doesn't specify. It's just asking about the condition that the angles must meet to create such a triangle. Sep 26, 2013 at 23:16

Hint: If you scale the triangle by a factor of $\alpha$, the perimeter is scaled by $\alpha$ and the area is scaled by $\alpha^2$.

Hence, given any triangle, there is a unique triangle that is similar to it, which satisfies perimeter = area.

• The question is about the property that the angles of all such triangles share. Sep 26, 2013 at 23:21
• @Jonn_Underwood There is no restriction (apart from being positive and summon to 180). Given any $a+b+c=180$, you can find a triangle with those angles, and perimeter = area. Sep 26, 2013 at 23:22
• See thats exactly what I thought and that's what's puzzling me, my professor posted this question as extra credit and I'm at a loss to what property he's talking about. Sep 26, 2013 at 23:24
• @Jonn_Underwood Ask him. He could have meant integer sides, or he could have wanted the scaling argument. Sep 26, 2013 at 23:25
• I did that, he said that's the whole point of the question, is to figure it out. Which is why I came to S.E. Sep 26, 2013 at 23:26

This a Pythagorean triple treatment of area to perimeter ratios where $$\angle AB$$ is right and $$A^2+B^2=C^2$$.

We can find Pythagorean triples, if they exist, for any ratio $$R$$ of area/perimeter by finding the $$m,n$$(s) that represent them using the following formula which includes a defined finite search for values of $$m$$. Whenever the $$m$$ $$R$$ combination yields a positive integer for $$n$$, we have the $$m,n$$ for a triple. We begin by solving the area/perimeter equation for $$n$$ where area=$$D$$ so as not to confuse it with $$A,B,C$$.

$$D=\frac{AB}{2}=\frac{(m^2-n^2)*2mn}{2}=mn(m^2-n^2)\quad P=(m^2-n^2)+2mn+(m^2+n^2)=2m^2+2mn$$

$$\frac{D}{P}=\frac{mn(m^2-n^2)}{2m^2+2mn}=\frac{mn(m-n)(m+n)}{2m(m+n)}=\frac{n(m-n)}{2}=R\qquad n^2-mn+2R=0$$

$$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{m\pm\sqrt{m^2-4*1*2R}}{2*1}$$

$$n=\frac{m\pm \sqrt{m^2-8R}}{2}\text{ where }\lceil\sqrt{8R}\space \rceil \le m \le 2R+1$$

Example: For $$R=0.5\quad \sqrt{8*0.5}=2\le m \le 2*0.5+1=2$$

$$n=\frac{2\pm \sqrt{2^2-8*0.5}}{2}=1\qquad f(2,1)=(3,4,5)$$ But you are asking about $$R=1$$ and there only two such Pythagorean triples to be found.

$$R=1\rightarrow 3\le m \le 3\quad f(3,2)=(5,12,13)\quad f(3,1)=(8,6,10)$$ For other ratios: $$R=1.5\rightarrow 4\le m \le 4\quad f(4,3)=(7,24,25)\quad f(4,1)=(15,8,17)$$

$$R=2\rightarrow 4\le m \le 5\quad f(4,2)=(12,16,20)\quad f(5,4)=(9,40,41)\quad f(5,1)=(24,10,26)$$

$$R=2.5\rightarrow 4\le m \le 6\quad f(6,5)=(11,60,61)\quad f(6,1)=(35,12,37)$$

$$R=3\rightarrow 4\le m \le 7\quad f(5,3)=(16,30,34)\quad f(5,2)=(21,20,29)\quad f(7,6)=(13,84,85)\quad f(7,1)=(48,14,50)$$

$$R=18\rightarrow 12\le m \le 37\quad f(12,6)=(108,144,180)\quad f(13,9)=(88,234,250)\quad f(13,4)=(153,104,185)\quad f(15,12)=(81,360,369)\quad f(15,3)=(216,90,234)\quad f(20,18)=(76,720,724)\quad f(20,2)=(396,80,404)\quad f(37,36)=(73,2664,2665)\quad f(37,1)=(1368,74,1370)$$

Aside: you can always find one primitive triple for a given $$R$$ if you let $$(m,n)=(2R+1,2R)$$.