Difference between revisions of "2023 AIME I Problems/Problem 13"
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such as the one shown below. | such as the one shown below. | ||
− | + | <asy> | |
+ | unitsize(2cm); | ||
+ | pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); | ||
+ | |||
+ | draw(o--u--(u+v)); | ||
+ | draw(o--v--(u+v), dotted); | ||
+ | draw(shift(w)*(o--u--(u+v)--v--cycle)); | ||
+ | draw(o--w); | ||
+ | draw(u--(u+w)); | ||
+ | draw(v--(v+w), dotted); | ||
+ | draw((u+v)--(u+v+w)); | ||
+ | </asy> | ||
==Solution 1 (3-D Vector Analysis)== | ==Solution 1 (3-D Vector Analysis)== | ||
Line 132: | Line 143: | ||
However, <math>ACD</math> and <math>A'C'D'</math> are both half the area of a rhombus with diagonals <math>\sqrt{31}</math> and <math>\sqrt{21}</math>, so our ratio is really | However, <math>ACD</math> and <math>A'C'D'</math> are both half the area of a rhombus with diagonals <math>\sqrt{31}</math> and <math>\sqrt{21}</math>, so our ratio is really | ||
− | < | + | <cmath>\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')}.</cmath> |
− | Because the diagonals of all of the faces are < | + | Because the diagonals of all of the faces are <math>\sqrt{31}</math> and <math>\sqrt{21}</math>, each edge of the parallelepipeds is <math>\sqrt{13}</math> by the Pythagorean theorem. |
− | We have < | + | We have <math>AB = AC = AD = \sqrt{13}</math>, and <math>BC = CD = BD = \sqrt{21}</math>. When we drop a perpendicular to the centroid of <math>BCD</math> from <math>A</math> (let's call this point <math>O</math>), we have <math>BO = \frac{\sqrt{21}}{\sqrt{3}} = \sqrt{7}</math>. Thus, |
<cmath>AB^2 - BO^2 = AO^2</cmath> | <cmath>AB^2 - BO^2 = AO^2</cmath> | ||
Line 142: | Line 153: | ||
<cmath>AO = \sqrt{6}.</cmath> | <cmath>AO = \sqrt{6}.</cmath> | ||
− | The area of base < | + | The area of base <math>BCD</math> is <math>\frac{21\sqrt{3}}{4}</math>. Hence, |
<cmath>\textrm{Vol}(ABCD) = \frac{\sqrt{6}\cdot\frac{21\sqrt{3}}{4}}{3}</cmath> | <cmath>\textrm{Vol}(ABCD) = \frac{\sqrt{6}\cdot\frac{21\sqrt{3}}{4}}{3}</cmath> | ||
<cmath> = \frac{63\sqrt{2}}{12}. </cmath> | <cmath> = \frac{63\sqrt{2}}{12}. </cmath> | ||
− | We can apply a similar approach to < | + | We can apply a similar approach to <math>A'B'C'D'</math>. |
− | < | + | <math>A'B' = A'C' = A'D' = \sqrt{13}</math>, and <math>B'C' = C'D' = B'D' = \sqrt{31}</math>. When we drop a perpendicular to the centroid of <math>B'C'D'</math> from <math>A'</math> (let's call this point <math>O'</math>), we have <math>B'O' = \frac{\sqrt{31}}{\sqrt{3}} = \sqrt{\frac{31}{3}}</math>. Thus, |
<cmath>A'B'^2 - B'O'^2 = A'O'^2</cmath> | <cmath>A'B'^2 - B'O'^2 = A'O'^2</cmath> | ||
Line 155: | Line 166: | ||
<cmath>A'O' = \sqrt{8}{3} = \frac{2\sqrt{6}}{3}.</cmath> | <cmath>A'O' = \sqrt{8}{3} = \frac{2\sqrt{6}}{3}.</cmath> | ||
− | The area of base < | + | The area of base <math>B'C'D'</math> is <math>\frac{31\sqrt{3}}{4}</math>. Hence, |
<cmath>\textrm{Vol}(A'B'C'D') = \frac{\frac{2\sqrt{6}}{3}\cdot\frac{31\sqrt{3}}{4}}{3}</cmath> | <cmath>\textrm{Vol}(A'B'C'D') = \frac{\frac{2\sqrt{6}}{3}\cdot\frac{31\sqrt{3}}{4}}{3}</cmath> | ||
Line 163: | Line 174: | ||
Finally, | Finally, | ||
− | <cmath>\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')} = \frac{\frac{63\sqrt{2}}{12}}{\frac{62\sqrt{2}}{12} = \frac{63}{62}.</cmath> | + | <cmath>\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')} = \frac{\frac{63\sqrt{2}}{12}}{\frac{62\sqrt{2}}{12}} = \frac{63}{62}.</cmath> |
− | Our answer is < | + | Our answer is <math>63 + 62 = \boxed{125}</math>. |
~mathboy100 | ~mathboy100 | ||
+ | |||
+ | ==Solution 4 (Pythagorean theorem)== | ||
+ | |||
+ | Since the two parallelepipeds have the same base, all we need to do is to find their respective heights. | ||
+ | |||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2; | ||
+ | |||
+ | label("A",a,S); | ||
+ | label("B",b,S); | ||
+ | label("C",c,S); | ||
+ | label("D",d,N); | ||
+ | label("P",p,S); | ||
+ | label("O",o,E); | ||
+ | |||
+ | draw(a--b--(b+c)); | ||
+ | draw(a--c--(b+c), dotted); | ||
+ | draw(shift(d)*(a--b--(b+c)--c--cycle)); | ||
+ | draw(a--d); | ||
+ | draw(b--(b+d)); | ||
+ | draw(c--(c+d), dotted); | ||
+ | draw((b+c)--(b+c+d)); | ||
+ | |||
+ | draw(d--p, dotted); | ||
+ | draw(c--b, dotted); | ||
+ | draw(a--(b+c), dotted); | ||
+ | draw(p--c, dotted); | ||
+ | draw(d--c, dotted); | ||
+ | </asy> | ||
+ | |||
+ | As illustrated in the above diagram, drop a perpendicular from <math>D</math> onto the base at <math>P</math>. Denote the center of the base by <math>O</math>. By symmetry, <math>P</math> must be on <math>AO</math>. Now we need to find <math>DP</math>. | ||
+ | |||
+ | Apply Pythagorean theorem to <math>\triangle DPA</math> we have | ||
+ | <cmath> DP^2 = AD^2 - AP^2.</cmath> | ||
+ | |||
+ | Apply Pythagorean theorem to <math>\triangle DPC</math> and then <math>\triangle CPO</math> we have | ||
+ | <cmath> DP^2 = DC^2 - CP^2 = DC^2 - (CO^2 + OP^2) = DC^2 - (CO^2 + (AO-AP)^2) = DC^2 - CO^2 - (AO-AP)^2.</cmath> | ||
+ | |||
+ | Combining the above two, we have | ||
+ | <cmath>AD^2 - AP^2 = DC^2 - CO^2 - (AO-AP)^2.</cmath> | ||
+ | |||
+ | Since <math>AD=\sqrt{13}</math>, <math>DC=\sqrt{21}</math>, <math>CO=\frac{\sqrt{21}}{2}</math>, <math>AO=\frac{\sqrt{31}}{2}</math>, plug them into the above equation and solve for the only unknown variable <math>AP</math>, we get <math>AP=\frac{5}{\sqrt{31}}.</math> | ||
+ | |||
+ | Thus the height | ||
+ | <cmath> DP = \sqrt{AD^2 - AP^2} = \sqrt{13 - \frac{25}{31}} = \sqrt{\frac{378}{31}}. </cmath> | ||
+ | |||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = 0.7*dir(80), e = c+d, p = 0.9*dir(10), o = (b+c)/2; | ||
+ | |||
+ | label("A'",a,S); | ||
+ | label("B'",b,S); | ||
+ | label("C'",c,W); | ||
+ | label("E'",e,N); | ||
+ | label("P'",p,S); | ||
+ | label("O'",o,W); | ||
+ | |||
+ | draw(a--b--(b+c)); | ||
+ | draw(a--c--(b+c), dotted); | ||
+ | draw(shift(d)*(a--b--(b+c)--c--cycle)); | ||
+ | draw(a--d); | ||
+ | draw(b--(b+d)); | ||
+ | draw(c--(c+d), dotted); | ||
+ | draw((b+c)--(b+c+d)); | ||
+ | |||
+ | draw(e--p, dotted); | ||
+ | draw(c--b, dotted); | ||
+ | draw(a--(b+c), dotted); | ||
+ | draw(p--a, dotted); | ||
+ | draw(e--a, dotted); | ||
+ | </asy> | ||
+ | |||
+ | For the other parallelepiped, using the same approach and drop a perpendicular from <math>E'</math> onto the base at <math>P'</math>. Similarly applying Pythagorean theorem to <math>\triangle E'P'C'</math>, <math>\triangle E'P'A'</math> and <math>\triangle A'P'O'</math> we have | ||
+ | <cmath>C'E'^2 - C'P'^2 = A'E'^2 - A'O'^2 - (C'P'-C'O')^2.</cmath> | ||
+ | |||
+ | Plugging known values into the above equation and solve for the only unknown variable <math>C'P'</math>, we get <math>C'P'=\frac{5}{\sqrt{21}}.</math> | ||
+ | |||
+ | Thus the height | ||
+ | <cmath> E'P' = \sqrt{C'E'^2 - C'P'^2} = \sqrt{13 - \frac{25}{21}} = \sqrt{\frac{248}{21}}. </cmath> | ||
+ | |||
+ | The ratio between the two is therefore | ||
+ | <cmath> \frac{DP}{E'P'} = \frac{\sqrt{\frac{378}{31}}}{\sqrt{\frac{248}{21}}} = \sqrt{\frac{2\cdot3^3\cdot7}{31}\cdot\frac{3\cdot7}{2^3\cdot31}} = \frac{3^2\cdot7}{2\cdot31} = \frac{63}{62}</cmath>, giving <math>\boxed{125}</math>. | ||
+ | |||
+ | ~sgdzw | ||
+ | |||
+ | ==Solution 5 (Visual)== | ||
+ | [[File:2023 AIME I 13a.png|350px|right]] | ||
+ | Let us inscribe a tetrahedron <math>ACB'D'</math> in given parallelepiped so that its edges coincide with the diagonals of the faces of the parallelepiped. Note that the three edges outgoing from the vertex <math>B'</math> have the same length <math>b</math>, and the three edges at the base have a different length <math>a.</math> | ||
+ | |||
+ | The volume of the tetrahedron <math>V= \frac {a^2}{12}\sqrt{3b^2 - a^2}</math> is three times less than the volume of the parallelepiped. | ||
+ | |||
+ | In second parallelepiped <math>a</math> and <math>b</math> change the positions. | ||
+ | |||
+ | Required ratio is <math>\frac {a^2 \cdot \sqrt {3b^2 - a^2}}{b^2 \cdot \sqrt {3a^2 - b^2}} = \frac {21 \sqrt {3 \cdot 31 - 21}}{31 \cdot \sqrt{3 \cdot 21 - 31}} = \frac {21 \sqrt {72}}{31 \sqrt {32}}= \frac {63}{62}.</math> | ||
+ | |||
+ | <i><b>Claim</b></i> | ||
+ | [[File:2023 AIME I 13.png|300px|right]] | ||
+ | Let <math>ABCD</math> be the regular pyramid, <math>AB = AC = AD = b, BC = BD = CD = a.</math> | ||
+ | |||
+ | The area of <math>\triangle BCD = \frac {a^2 \sqrt {3}}{4}.</math> | ||
+ | |||
+ | Height <math>AO^2 = AB^2 - OB^2 = {b^2 - \frac {a^2}{3}}.</math> | ||
+ | |||
+ | Volume <math>V= \frac {a^2}{12}\sqrt{3b^2 - a^2}.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2, x = (0.2, 0); | ||
+ | |||
+ | label("A",a,S); | ||
+ | label("B",b,S); | ||
+ | label("C",c,S); | ||
+ | label("D",d,N); | ||
+ | label("X",x,S); | ||
+ | |||
+ | draw(a--b--(b+c)); | ||
+ | draw(a--c--(b+c), dotted); | ||
+ | draw(shift(d)*(a--b--(b+c)--c--cycle)); | ||
+ | draw(a--d); | ||
+ | draw(b--(b+d)); | ||
+ | draw(d--x); | ||
+ | draw(x--c, dotted); | ||
+ | draw(c--(c+d), dotted); | ||
+ | draw((b+c)--(b+c+d)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | Let <math>\angle CAB</math> be acute and let <math>X</math> be the foot of the altitude from <math>C</math> to <math>AB</math>. Notice that this also implies that <math>X</math> is the foot of the altitude from <math>D</math> to <math>AB</math>. Now <math>\sqrt{13} \cdot CX = AB \cdot CX = \frac{1}{2} \sqrt{21} \cdot \sqrt{31}</math> so <math>CX = \frac{\sqrt{21} \cdot \sqrt{31}}{2 \sqrt{13}}</math> and <math>DX</math> is the same. <math>CD</math> must either be <math>\sqrt{21}</math> or <math>\sqrt{31}</math> because it is a diagonal. If <math>CD = \sqrt{21}</math>, applying the Law of Cosines on <math>\angle DXC</math>, <math>\cos \angle DXC = -\frac{5}{21}</math> so <math>\sin \angle DXC = \sqrt{\frac{416}{441}}</math>. If <math>CD = \sqrt{31}</math>, <math>\cos \angle DXC = \frac{5}{31}</math>, so <math>\sin \angle DXC = \sqrt{\frac{936}{961}}</math>. The ratios of the two parallelepipeds is equal to the ratios of the heights (since the bases are the same) which is equal to the ratio of the sines. Therefore it is <cmath> \frac{\sqrt{\frac{936}{961}}}{\sqrt{\frac{416}{441}}} = \sqrt{\frac{936}{416}} \cdot \frac{21}{31} = \frac{3}{2} \cdot \frac{21}{31} = \frac{63}{62} </cmath> so the answer is <math>63 + 62 = \boxed{125}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/5mJ6EqdFD94 | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | |||
+ | |||
+ | ==Animated Video Solution== | ||
+ | |||
+ | https://youtu.be/VvCl5KIqT9M | ||
+ | |||
+ | ~Star League (https://starleague.us) | ||
{{AIME box|year=2023|n=I|num-b=12|num-a=14}} | {{AIME box|year=2023|n=I|num-b=12|num-a=14}} |
Latest revision as of 20:25, 5 September 2024
Contents
Problem
Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths and . The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is , where and are relatively prime positive integers. Find . A parallelepiped is a solid with six parallelogram faces such as the one shown below.
Solution 1 (3-D Vector Analysis)
Denote . Denote by the length of each side of a rhombus.
Now, we put the solid to the 3-d coordinate space. We put the bottom face on the plane. For this bottom face, we put a vertex with an acute angle at the origin, denoted as . For two edges that are on the bottom face and meet at , we put one edge on the positive side of the -axis. The endpoint is denoted as . Hence, . We put the other edge in the first quadrant of the plane. The endpoint is denoted as . Hence, .
For the third edge that has one endpoint , we denote by its second endpoint. We denote . Without loss of generality, we set . Hence,
We have and
Case 1: or .
By solving (2) and (3), we get
Plugging these into (1), we get
Case 2: and , or and .
By solving (2) and (3), we get
Plugging these into (1), we get
We notice that . Thus, (4) (resp. (5)) is the parallelepiped with a larger (resp. smaller) height.
Therefore, the ratio of the volume of the larger parallelepiped to the smaller one is
Recall that . Thus, . Plugging this into the equation above, we get
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (no trig)
Let one of the vertices be at the origin and the three adjacent vertices be , , and . For one of the parallelepipeds, the three diagonals involving the origin have length . Hence, and . Since all of , , and have equal length, , , and . Symmetrically, , , and . Hence the volume of the parallelepiped is given by .
For the other parallelepiped, the three diagonals involving the origin are of length and the volume is .
Consequently, the answer is , giving .
~EVIN-
Solution 3 (No trig, no linear algebra)
Observe that both parallelepipeds have two vertices (one on each base) that have three congruent angles meeting at them. Denote the parallelepiped with three acute angles meeting , and the one with three obtuse angles meeting .
The area of a parallelepiped is simply the base area times the height, but because both parallelepipeds have the same base, what we want is just the ratio of the heights.
Denote the point with three acute angles meeting at it in as , and its neighbors , , and . Similarly, denote the point with three obtuse angles meeting at it in as , and its neighbors , , and .
We have the following equations:
However, and are both half the area of a rhombus with diagonals and , so our ratio is really
Because the diagonals of all of the faces are and , each edge of the parallelepipeds is by the Pythagorean theorem.
We have , and . When we drop a perpendicular to the centroid of from (let's call this point ), we have . Thus,
The area of base is . Hence,
We can apply a similar approach to .
, and . When we drop a perpendicular to the centroid of from (let's call this point ), we have . Thus,
The area of base is . Hence,
Finally,
Our answer is .
~mathboy100
Solution 4 (Pythagorean theorem)
Since the two parallelepipeds have the same base, all we need to do is to find their respective heights.
As illustrated in the above diagram, drop a perpendicular from onto the base at . Denote the center of the base by . By symmetry, must be on . Now we need to find .
Apply Pythagorean theorem to we have
Apply Pythagorean theorem to and then we have
Combining the above two, we have
Since , , , , plug them into the above equation and solve for the only unknown variable , we get
Thus the height
For the other parallelepiped, using the same approach and drop a perpendicular from onto the base at . Similarly applying Pythagorean theorem to , and we have
Plugging known values into the above equation and solve for the only unknown variable , we get
Thus the height
The ratio between the two is therefore , giving .
~sgdzw
Solution 5 (Visual)
Let us inscribe a tetrahedron in given parallelepiped so that its edges coincide with the diagonals of the faces of the parallelepiped. Note that the three edges outgoing from the vertex have the same length , and the three edges at the base have a different length
The volume of the tetrahedron is three times less than the volume of the parallelepiped.
In second parallelepiped and change the positions.
Required ratio is
Claim
Let be the regular pyramid,
The area of
Height
Volume
vladimir.shelomovskii@gmail.com, vvsss
Solution 6
Let be acute and let be the foot of the altitude from to . Notice that this also implies that is the foot of the altitude from to . Now so and is the same. must either be or because it is a diagonal. If , applying the Law of Cosines on , so . If , , so . The ratios of the two parallelepipeds is equal to the ratios of the heights (since the bases are the same) which is equal to the ratio of the sines. Therefore it is so the answer is .
Video Solution
~MathProblemSolvingSkills.com
Animated Video Solution
~Star League (https://starleague.us)
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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