Perimeter 2008

Let the three sides be a, a and b. Hence 2a + b = 2008 - equation 1

Using the triangular inequality we have 2a > b - equation 2

Also using the cosine rule for finding sides of a triangle we note that where is the angle between the two equal sides and obtuse (none of the equal angles can be obtuse as a triangle cannot have more than one obtuse angle). Solving the inequalities we get 502 < a < 588.25 allowing total 86 values of a. Thus there are 86 such triangles.

Copy of my solution of the same problem .... "All I'm given is the Perimeter?" by Mridul Jain. $For~ a~ right~ angled~ triangle,~ with~~ legs=x,~ we~ have ~Hypothesis,~h~=x\sqrt2.\\ So~x=\dfrac{2008}{2+\sqrt2}=588.12...and~h=831.55...\\ So ~upper~ limit~for~x ~is ~588~~and~~for~base~b~lower~limit~is~832>h.\\ If~all~three~are~in~line,~we~have~502-502-1004.\\ So~ b=1003~would~ be~ OK. But~x~is~not~an~integer.\\ So~b=1002~is~higher~limit~and~x=503~lower~limit.\\ Number~of~solutions-moving~ in~ steps~ of~2~to~keep~x~as~an~integer~is~\dfrac{1002-832} 2+1=86.\\ Check~for~x~Number~of~solutions-moving~ in ~steps~ of~1~~is~~(588 -503)+1=\Large~~\color{#D61F06}{86,~checked.}$