Another solution: Be AB = c, s = (a + b + c)/2 = 50, h the height of EDC on DE, r the radius of the incircle and A the area of ABC. Then, the height of ABC on c is 2r + h. As ABC and EDC are similar, so we get c / 12 = (2r +h) / h. Solving for h, we get h = 24r / (c -12). The Area A is given by A = sr = 50r and also by A = c(2r + h) / 2. Equating these two, we get 50r = c(2r + h) / 2. Inserting the expression for h we got from the similarity condition, we have the equation 50r = c(2r + 24r / (c -12)) / 2. Cancelling r and simplifying, the result is the quadratic equation c² - 50c + 600 = 0 with solutions c = 20 and c = 30.

Draw a perpendicular from C to DE which intersects at F and let y = CF and let r be the radius of the circle and a = AB. Area(Trapezoid(ADEB)) = (a + 12) * r and A(triangle(CDE)) = 6 * y implies Area(Triangle(ABC)) = 50 * r = (a + 12) * r + 6 * y. Triangle(ABC) ~ Triangle(CDE) implies a/12 = (2 * r + y)/y implies y = 24 * r/(a - 12) implies 50 * r = (a + 12) * r + 6 * (24 * r/(a - 12)) implies 50 * (a - 12) * r = a^2 * r implies a^2 - 50 * a + 600 = 0 implies (a - 30) * (a - 20) = 0 implies a = 30 and a = 20. We obtain: a = 30 implies b + c = 70 and a = 20 implies b + c = 80. Note: For c arbitrary we have (30, 70 - c, c) and (20, 80 - c, c). Also for a = 30 implies r = sqrt(10 * (c - 20) * (50 - c))/5, where (c > 20) and for a = 20 implies r = sqrt(15 * (c - 30) * (50 - c))/5, where (c > 30).

If the radius of the circle is r AB=x and the height of △ABC is h, then the height of △CDE is 12(2r/(x-12))=24r/(x-12), so h=2r+24r/(x-12)) ① Area of △ABC=(1/2)xh=(1/2)r100 ∴xh=100r ② From ①②, x(2r+24r/(x-12))=100r ∴x^2-50x+600=(x -20)(x-30)=0 ∴x=20, 30

Let the tangents from A ,B and C be a , b and c respectively. The perimeter of triangle ABC = 2a +2b +2c = 100 Hence a+b+c=50 Perimeter of triangle CDE = 2c Triangles CDE and CAB are similar. ( a a )Scale factor would be a+b/12 2c •a+b/12=100 c(50-c)=600 Solve for c would get c= 20 or30 Hence a+b= 30 or20 Hence AB= 30 or 20

WLOG, ABC is isosceles. CE/DE = CB/AB. One minute exercise.

Very beautiful problem. Whether 20 or 30 both are possible ?

Can't see yet how the circle fits there two ways... it must be possible though. Angle ACB variating.

Why 2 solutions?

🔺️menor Lados x; y; 12 Perímetro P1 = x + y + 12 Trapézio Lados a =?; b; c; d = 12 Perímetro P2=a+b+c+d Teorema de Pitot P2 = 2a + 24 🔺grande Perímetro P3 = P1+P2-24 P3 = P1 + 2a = 100 Semelhança de 🔺️ P1/P3 = 12/a P1 = 1200/a 1200/a + 2a = 100 *a1 = 30* *a2 = 20*

Very nice job

this is stuff enough for a nested calculation: 10 lde=12:lu=100:lh=1:r=lde/3:sw=.1:dim x(2),y(2):goto 400 20 goto 60 30 hg=2*r*lab/(lab-lde) 40 dfu1=sqr(lh^2+hg^2)/lu:dfu2=sqr((lab-lh)^2+hg^2)/lu:dfu3=lab/lu:df=dfu1+dfu2+dfu3-1 50 return 60 lab=lu/8:gosub 30 70 df1=df:lab1=lab:lab=lab+sw: if hg>100*lde then stop:rem r=r-sw:goto 50 80 lab2=lab:gosub 30:if df1*df>0 then 70 90 lab=(lab1+lab2)/2:gosub 30:if df1*df>0 then lab1=lab else lab2=lab 100 rem zi=zi+1:if zi>100 then r=r-sw:goto 50 110 if abs(df)>1E-10 then 90 120 x(0)=0:y(0)=0:x(1)=lab:y(1)=0:x(2)=lh:y(2)=hg 130 print hg,"%",lab,"%",r:rem den mittelpunkt berechnen ** 140 dx=lh:dy=hg:nd=sqr(hg^2+lh^2):xg11=dy/nd*r:yg11=-dx/nd*r:xg12=xg11+dx:yg12=yg11+dy 150 xg21=0:yg21=r:xg22=1:yg22=yg21:gosub 160:xm1=xl:ym1=yl:goto 230 160 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12) 170 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22) 180 a13=a131+a132:a23=a231+a232: ngl1=a12*a21:ngl2=a22*a11 190 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end 200 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 210 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 220 xl=zx/ngl:yl=zy/ngl:print "x=";xl;"y=";yl :return 230 dx=lab-lh:dy=-hg:nd=sqr(dx^2+dy^2) 240 xg11=lab+r*dy/nd:yg11=-r*dx/nd:xg12=xg11+dx:yg12=yg11+dy 250 gosub 160:xm2=xl:ym2=yl:mass=1000/hg:if mass>1000/lh then mass=1000/lh 255 goto 270 260 xb=x*mass:yb=y*mass:return 270 x=0:y=0:gosub 260:xba=xb:yba=yb:for a=1 to 3:ia=a:if a=3 then ia=0 280 x=x(ia):y=y(ia):gosub 260:xbn=xb:ybn=yb:gosub 290:goto 300 290 line xba,yba,xbn,ybn:return 300 xba=xbn:yba=ybn:next a:x=xm1:y=ym1:rem gosub 250:circle xb,yb,r*mass 310 x=xm2:y=ym2:rem gosub 250:circle xb,yb,r*mass 320 xg11=0:yg11=0:xg12=xm1:yg12=ym1 330 xg21=lab:yg21=0:xg22=xm2:yg22=ym2:gosub 160:x=xl:y=yl:gosub 260:rn=yl 340 xmb=xb:ymb=yb:gcol 9:circle xmb,ymb, rn*mass 350 xg11=0:yg11=2*rn:xg12=1:yg12=yg11:xg21=0:yg21=0:xg22=lh:yg22=hg:gosub 160:xs1=xl:ys1=yl 360 xg21=lh:yg21=hg:xg22=lab:yg22=0:gosub 160:xs2=xl:ys2=yl 370 lder=sqr((xs1-xs2)^2+(ys2-ys1)^2):dg=(lde-lder)/lde 380 x=xs1:y=ys1:gosub 260:xba=xb:yba=yb:x=xs2:y=ys2:gosub 260:xbn=xb:ybn=yb 390 gcol 8:gosub 290:return 400 gosub 20 410 cls:lh1=lh:dg1=dg:lh=lh+sw/100:lh2=lh:gosub 20:if dg1*dg>0 then 410 420 lh=(lh1+lh2)/2:gosub 20:if dg1*dg>0 then lh1=lh else lh2=lh 430 if abs(dg)>1E-10 then 420 440 cls:print rn,"%",lab:gosub 20 4% 20 20% 20% 4 x=18.5634884y=4 x=18.5634884y=4 x=18.5634884y=4 x=17.7015814y=8 x=29.7015814y=8 > run in bbc basic sdl, hit ctrl tab to copy.

How is it possible to get two solutions for one...

No way…