Two touching balls with radii $a$ and $b$ are enclosed in a cylindrical tin of diameter $d$ . Both balls hit the top surface and the shell of the cylinder. The largest ball also hits the bottom surface. Show that $\sqrt{d} =\sqrt{a} +\sqrt{b}$
Problem
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Tags: geometry, 3D geometry, sphere, cylinder
mathverse06
08.08.2020 13:08
One of the reasons why changing 3D to 2D is important. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.36, xmax = 15.36, ymin = -7.04, ymax = 7.04; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen ffdxqq = rgb(1,0.8431372549019608,0); draw((-7.2,3.88)--(-7.22,-1.84)--(0.44,-1.8)--(0.34,3.94)--cycle, linewidth(1.2)); draw(circle((-4.6,0.84), 2.616907150615757), linewidth(2) + ccqqqq); draw(circle((-1,2.56), 1.3730906283528068), linewidth(2) + ffdxqq); /* draw figures */ draw((-7.2,3.88)--(-7.22,-1.84), linewidth(1.2)); draw((-7.22,-1.84)--(0.44,-1.8), linewidth(1.2)); draw((0.44,-1.8)--(0.34,3.94), linewidth(1.2)); draw((0.34,3.94)--(-7.2,3.88), linewidth(1.2)); /* dots and labels */ dot((-7.2,3.88),dotstyle); label("$A$", (-7.12,4.08), NE * labelscalefactor); dot((-7.22,-1.84),dotstyle); label("$B$", (-7.14,-1.64), NE * labelscalefactor); dot((0.44,-1.8),dotstyle); label("$C$", (0.52,-1.6), NE * labelscalefactor); dot((0.34,3.94),dotstyle); label("$D$", (0.42,4.14), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] we get $a+b+2\sqrt{ab}=d$ Therefore, $\sqrt{d} =\sqrt{a} +\sqrt{b}$. Kinda went fast. Sorry about that. PS: I drew the wrong diagram by mistake the large circle needs to touch the top.