Difference between revisions of "2015 AIME I Problems/Problem 15"
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A block of wood has the shape of a right circular cylinder with radius <math>6</math> and height <math>8</math>, and its entire surface has been painted blue. Points <math>A</math> and <math>B</math> are chosen on the edge of one of the circular faces of the cylinder so that <math>\overarc{AB}</math> on that face measures <math>120^\text{o}</math>. The block is then sliced in half along the plane that passes through point <math>A</math>, point <math>B</math>, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is <math>a\cdot\pi + b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>. | A block of wood has the shape of a right circular cylinder with radius <math>6</math> and height <math>8</math>, and its entire surface has been painted blue. Points <math>A</math> and <math>B</math> are chosen on the edge of one of the circular faces of the cylinder so that <math>\overarc{AB}</math> on that face measures <math>120^\text{o}</math>. The block is then sliced in half along the plane that passes through point <math>A</math>, point <math>B</math>, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is <math>a\cdot\pi + b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>. | ||
+ | |||
+ | <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 | ||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtube.com/watch?v=WkSksszz4KM | ||
==Solution== | ==Solution== | ||
+ | Label the points where the plane intersects the top face of the cylinder as <math>C</math> and <math>D</math>, and the center of the cylinder as <math>O</math>, such that <math>C,O,</math> and <math>A</math> are collinear. Let <math>T</math> be the center of the bottom face, and <math>M</math> the midpoint of <math>\overline{AB}</math>. Then <math>OT=4</math>, <math>TM=3</math> (because of the 120 degree angle), and so <math>OM=5</math>. | ||
+ | |||
+ | Project <math>C</math> and <math>D</math> onto the bottom face to get <math>X</math> and <math>Y</math>, respectively. Then the section <math>ABCD</math> (whose area we need to find), is a stretching of the section <math>ABXY</math> on the bottom face. The ratio of stretching is <math>\frac{OM}{TM}=\frac{5}{3}</math>, 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 <math>ABXY</math> is <math>18\sqrt{3}\ + 12 \pi</math>. Thus, the area of section <math>ABCD</math> is <math>20\pi + 30\sqrt{3}</math>, and so our answer is <math>20+30+3=\boxed{053}</math>. | ||
+ | |||
+ | ==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 <math>O</math> has become a circle with <math>\overarc{AB}</math> = <math>\overarc{CD}</math> = <math>120^\text{o}</math>. Using 30-60-90 triangles, we get rectangle <math>ABCD</math> to have a horizontal component of <math>6</math>. Now, consider a side view, such that <math>A</math> and <math>B</math> coincide at the bottom of the diagram. From this view, consider the right triangle composed of hypotenuse <math>AD</math> and a point along the base of the viewpoint, which will be labeled as <math>E</math>. From the top view, <math>AE = 6</math>. Because of the height of the cylinder, <math>DE</math> is equal to <math>8</math>. This makes <math>AD</math> equal to <math>10</math>. | ||
+ | |||
+ | |||
+ | Now, the use of simple calculus is required. Conceptualize an infinite number of lines perpendicular to <math>AE</math> intersecting both <math>AE</math> and <math>AD</math>. Consider the area between point <math>A</math> 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 <math>w</math>. The area of the base within these two positions approaches a rectangle with length AE' and width <math>w</math>. 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<math>w</math>:6<math>w</math> = 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 <math>18\sqrt{3}\ + 12 \pi</math>. The area of the unpainted surface therefore becomes <math>20\pi + 30\sqrt{3}</math>, and so our answer is <math>\boxed{053}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | This problem can be calculus-bashed (for those like me who never noticed the surface was merely a stretch of its projection). Label points as in the first paragraph of Solution 1 (<math>A</math> and <math>B</math> as given, <math>M</math> the midpoint of <math>AB</math>, <math>O</math> the center of the cylinder, <math>T</math> the center of the bottom face of the cylinder). Because of the 120 degrees and right triangle calculations, we find <math>MT</math> = 3, <math>OT</math> = 4, <math>OM</math> = 5). We will be integrating with respect to the y-coordinate which we define as distance downwards from <math>O</math> (in this system, the <math>y</math>-coordinate of the bottom face would be 4). | ||
+ | |||
+ | We note that by similar triangles, we have that the length from <math>O</math> to the point on the unpainted surface of coordinate <math>y</math> is <math>\ell = \frac{5}{4} y</math>, and therefore <math>d\ell = \frac{5}{4} dy</math>. Define the segment <math>A'B'</math> to be the intersection of the painted surface with the circular cross section of the cylinder of coordinate <math>y</math>, with endpoints <math>A'</math> and <math>B'</math> and midpoint <math>M'</math>, with <math>T'</math> the center of this circular cross section. Then, by similar triangles, <math>T'M' = \frac{3}{4} y</math> and thus <cmath>A'B' = 2A' M' = 2 \sqrt{ 6^2 - \left( \frac{3}{4}y\right)^2 } = \frac{3}{2} \sqrt{ 64 - y^2 }</cmath>. We know that <math>A'B'</math> is perpendicular to <math>OM</math>. | ||
+ | |||
+ | Now we can set up our integral: we will integrate <math>y</math> from 0 to 4 and multiply by two because the total height is 8. | ||
+ | <cmath> A = 2\int_0^4 \left(\frac{3}{2}\sqrt{ 64 - y^2 }\right) \left( \frac{5}{4} dy\right) </cmath> | ||
+ | <cmath> A = \frac{15}{4} \int_0^4 \sqrt{ 64 - y^2 }dy </cmath> | ||
− | + | Then we substitute <math>y = 8\sin \theta</math> with <math>dy = 8 \cos \theta d \theta</math>, changing the bounds to 0 to <math>\frac{\pi}{6}</math> as appropriate. | |
+ | <cmath> A = \frac{15}{4} \int_0^\frac{\pi}{6} \sqrt{ 64 - 64\sin^2 \theta }\cdot 8\cos\theta d\theta </cmath> | ||
+ | <cmath> A = 240 \int_0^\frac{\pi}{6} \cos^2 \theta d\theta </cmath> | ||
+ | <cmath> A = 240 \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_0^\frac{\pi}{6} = 240 \left[ \frac{\pi}{12} + \frac{\sqrt{3}}{8} \right] = 20{\pi} + 30\sqrt{3}</cmath> | ||
− | + | Therefore, <math>a + b + c = 20 + 30 + 3 = \boxed{053}</math>. | |
== See also == | == See also == | ||
{{AIME box|year=2015|n=I|num-b=14|after=Last Problem}} | {{AIME box|year=2015|n=I|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
− | [[Category: | + | |
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Revision as of 02:40, 9 March 2021
Contents
Problem
A block of wood has the shape of a right circular cylinder with radius and height , and its entire surface has been painted blue. Points and are chosen on the edge of one of the circular faces of the cylinder so that on that face measures . The block is then sliced in half along the plane that passes through point , point , and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is , where , , and are integers and is not divisible by the square of any prime. Find .
Credit to Royalreter1 and chezbgone2 for the diagram
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=WkSksszz4KM
Solution
Label the points where the plane intersects the top face of the cylinder as and , and the center of the cylinder as , such that and are collinear. Let be the center of the bottom face, and the midpoint of . Then , (because of the 120 degree angle), and so .
Project and onto the bottom face to get and , respectively. Then the section (whose area we need to find), is a stretching of the section on the bottom face. The ratio of stretching is , 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 is . Thus, the area of section is , and so our answer is .
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 has become a circle with = = . Using 30-60-90 triangles, we get rectangle to have a horizontal component of . Now, consider a side view, such that and coincide at the bottom of the diagram. From this view, consider the right triangle composed of hypotenuse and a point along the base of the viewpoint, which will be labeled as . From the top view, . Because of the height of the cylinder, is equal to . This makes equal to .
Now, the use of simple calculus is required. Conceptualize an infinite number of lines perpendicular to intersecting both and . Consider the area between point 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 . The area of the base within these two positions approaches a rectangle with length AE' and width . 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 10:6 = 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 . The area of the unpainted surface therefore becomes , and so our answer is .
Solution 3
This problem can be calculus-bashed (for those like me who never noticed the surface was merely a stretch of its projection). Label points as in the first paragraph of Solution 1 ( and as given, the midpoint of , the center of the cylinder, the center of the bottom face of the cylinder). Because of the 120 degrees and right triangle calculations, we find = 3, = 4, = 5). We will be integrating with respect to the y-coordinate which we define as distance downwards from (in this system, the -coordinate of the bottom face would be 4).
We note that by similar triangles, we have that the length from to the point on the unpainted surface of coordinate is , and therefore . Define the segment to be the intersection of the painted surface with the circular cross section of the cylinder of coordinate , with endpoints and and midpoint , with the center of this circular cross section. Then, by similar triangles, and thus . We know that is perpendicular to .
Now we can set up our integral: we will integrate from 0 to 4 and multiply by two because the total height is 8.
Then we substitute with , changing the bounds to 0 to as appropriate.
Therefore, .
See also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.