2015 AIME I Problems/Problem 15

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A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\text{o}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c}$, where $a$, $b$, and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$.

[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3);  picture lpic, rpic;  size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8));  size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,"$A$",(-3,3*sqrt(3),0),W); label(rpic,"$B$",(-3,-3*sqrt(3),0),W);  add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]

Credit to Royalreter1 and chezbgone2 for the diagram


Label the points where the plane intersects the top face of the cylinder as $C$ and $D$, and the center of the cylinder as $O$, such that $C,O,$ and $A$ are collinear. Let $T$ be the center of the bottom face, and $M$ the midpoint of $\overline{AB}$. Then $OT=4$, $TM=3$ (because of the 120 degree angle), and so $OM=5$.

Project $C$ and $D$ onto the bottom face to get $X$ and $Y$, respectively. Then the section $ABCD$ (whose area we need to find), is a stretching of the section $ABXY$ on the bottom face. The ratio of stretching is $\frac{OM}{TM}=\frac{5}{3}$, and we do not square this value when finding the area because it is only stretching in one direction. Using 30-60-90 triangles and circular sectors, we find that the area of the section $ABXY$ is $18\sqrt{3}\ + 12 \pi$. Thus, the area of section $ABCD$ is $20\pi + 30\sqrt{3}$, and so our answer is $20+30+3=\boxed{053}$.

Diagram: http://data.artofproblemsolving.com/images/wiki/c/ca/2015_AIME_I_P15_diagram_1.jpeg

Solution 2

Label the points same as in the first sentence above. Consider a view of the cylinder such that height is disregarded, i.e. a top view. From this view, note that Cylinder $O$ has become a circle with $\overarc{AB}$ = $\overarc{CD}$ = $120^\text{o}$. Using 30-60-90 triangles, we get rectangle $ABCD$ to have a horizontal component of $6$. Now, consider a side view, such that $A$ and $B$ coincide at the bottom of the diagram. From this view, consider the right triangle composed of hypotenuse $AD$ and a point along the base of the viewpoint, which will be labeled as $E$. From the top view, $AE = 6$. Because of the height of the cylinder, $DE$ is equal to $8$. This makes $AD$ equal to $10$.

Now, the use of simple calculus is required. Conceptualize an infinite number of lines perpendicular to $AE$ intersecting both $AE$ and $AD$. Consider the area between point $A$ and the first vertical line. Label the point where the line intersects AE as E', and the point where the line intersects AD as D'. The area of the part of the initial unpainted face within these two positions approaches a rectangle with length AD' and width $w$. The area of the base within these two positions approaches a rectangle with length AE' and width $w$. The ratio of AD':AE' is 10:6, since the ratio of AD:AE is 10:6. This means that the area of the initial unpainted surface within these two positions to the area of the base within these two positions is equal to 10$w$:6$w$ = 10:6. Through a similar argument, the areas between each set of vertical lines also maintains a ratio of 10:6. Therefore, the ratio of the area we wish to find to the area of the base between AB and CD (from the top perspective) is 10:6. Using 30-60-90 triangles and partial circles, the area of the base between AB and CD is calculated to be $18\sqrt{3}\ + 12 \pi$. The area of the unpainted surface therefore becomes $20\pi + 30\sqrt{3}$, and so our answer is $\boxed{053}$.

See also

2015 AIME I (ProblemsAnswer KeyResources)
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Problem 14
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