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| − | ==Problem==
| + | Sunwin hay còn được gọi thân thiện là sun20 được người chơi tôn vinh là game bài đổi thưởng uy tín số 1 châu Á. Thành viên được chơi các trò chơi đánh bài độc quyền, hoặc các trò tài xỉu MD5, Tài xỉu livestream có gái xinh ngồi trò chuyện real time. Ngoài ra là các trò chơi nổ hũ, tài xỉu, slot quay thưởng, live casino đặc sắc. Người chơi có thể chơi nhanh bản web hoặc tải sunwin về điện thoại mà không lo bị chặn. |
| − | A <i>regular octahedron</i> has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of <math>Q</math>?
| + | Địa chỉ: 19 P. Nguyễn Văn Giáp, Mễ Trì, Từ Liêm, Hà Nội, Việt Nam |
| − | | + | Website: https://trochoiviet.com.vn/ |
| − | <asy>
| |
| − | // Diagram by TheMathGuyd
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| − | import graph;
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| − | // The Solid
| |
| − | // To save processing time, do not use three (dimensions)
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| − | // Project (roughly) to two
| |
| − | size(15cm);
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| − | pair Fr, Lf, Rt, Tp, Bt, Bk;
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| − | Lf=(0,0);
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| − | Rt=(12,1);
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| − | Fr=(7,-1);
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| − | Bk=(5,2);
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| − | Tp=(6,6.7);
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| − | Bt=(6,-5.2);
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| − | draw(Lf--Fr--Rt);
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| − | draw(Lf--Tp--Rt);
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| − | draw(Lf--Bt--Rt);
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| − | draw(Tp--Fr--Bt);
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| − | draw(Lf--Bk--Rt,dashed);
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| − | draw(Tp--Bk--Bt,dashed);
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| − | label(rotate(-8.13010235)*slant(0.1)*"$Q$", (4.2,1.6));
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| − | label(rotate(21.8014095)*slant(-0.2)*"$?$", (8.5,2.05));
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| − | pair g = (-8,0); // Define Gap transform
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| − | real a = 8; | |
| − | draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow
| |
| − | // Time for the NET
| |
| − | pair DA,DB,DC,CD,O;
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| − | DA = (4*sqrt(3),0);
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| − | DB = (2*sqrt(3),6);
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| − | DC = (DA+DB)/3;
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| − | CD = conj(DC);
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| − | O=(0,0);
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| − | transform trf=shift(3g+(0,3));
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| − | path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB);
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| − | draw(trf*NET);
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| − | label("$7$",trf*DC);
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| − | label("$Q$",trf*DC+DA-DB);
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| − | label("$5$",trf*DC-DB);
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| − | label("$3$",trf*DC-DA-DB);
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| − | label("$6$",trf*CD);
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| − | label("$4$",trf*CD-DA);
| |
| − | label("$2$",trf*CD-DA-DB);
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| − | label("$1$",trf*CD-2DA);
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| − | </asy>
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| − | | |
| − | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math>
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| − | | |
| − | ==Solution 1==
| |
| − | We color face <math>6</math> red and face <math>5</math> yellow. Note that from the octahedron, face <math>5</math> and face <math>?</math> do not share anything in common. From the net, face <math>5</math> shares at least one vertex with all other faces except face <math>1,</math> which is shown in green:
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| − | <asy>
| |
| − | /*
| |
| − | Diagram by TheMathGuyd
| |
| − | Edited by MRENTHUSIASM
| |
| − | */
| |
| − | import graph;
| |
| − | // The Solid
| |
| − | // To save processing time, do not use three (dimensions)
| |
| − | // Project (roughly) to two
| |
| − | size(15cm);
| |
| − | pair Fr, Lf, Rt, Tp, Bt, Bk;
| |
| − | Lf=(0,0);
| |
| − | Rt=(12,1);
| |
| − | Fr=(7,-1);
| |
| − | Bk=(5,2);
| |
| − | Tp=(6,6.7);
| |
| − | Bt=(6,-5.2);
| |
| − | fill(Tp--Bk--Lf--cycle,red);
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| − | fill(Bt--Bk--Lf--cycle,yellow);
| |
| − | fill(Fr--Rt--Tp--cycle,green);
| |
| − | draw(Lf--Fr--Rt);
| |
| − | draw(Lf--Tp--Rt);
| |
| − | draw(Lf--Bt--Rt);
| |
| − | draw(Tp--Fr--Bt);
| |
| − | draw(Lf--Bk--Rt,dashed);
| |
| − | draw(Tp--Bk--Bt,dashed);
| |
| − | label(rotate(-8.13010235)*slant(0.1)*"$Q$", (4.2,1.6));
| |
| − | label(rotate(21.8014095)*slant(-0.2)*"$?$", (8.5,2.05));
| |
| − | pair g = (-8,0); // Define Gap transform
| |
| − | real a = 8;
| |
| − | draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow
| |
| − | // Time for the NET
| |
| − | pair DA,DB,DC,CD,O;
| |
| − | DA = (4*sqrt(3),0);
| |
| − | DB = (2*sqrt(3),6);
| |
| − | DC = (DA+DB)/3;
| |
| − | CD = conj(DC);
| |
| − | O=(0,0);
| |
| − | transform trf=shift(3g+(0,3));
| |
| − | path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB);
| |
| − | fill(trf*((DA-DB)--O--DA--cycle),red);
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| − | fill(trf*((DA-DB)--O--(-DB)--cycle),yellow);
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| − | fill(trf*((-2*DA)--(-DA-DB)--(-DA)--cycle),green);
| |
| − | draw(trf*NET);
| |
| − | label("$7$",trf*DC);
| |
| − | label("$Q$",trf*DC+DA-DB);
| |
| − | label("$5$",trf*DC-DB);
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| − | label("$3$",trf*DC-DA-DB);
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| − | label("$6$",trf*CD);
| |
| − | label("$4$",trf*CD-DA);
| |
| − | label("$2$",trf*CD-DA-DB);
| |
| − | label("$1$",trf*CD-2DA);
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| − | </asy>
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| − | Therefore, the answer is <math>\boxed{\textbf{(A)}\ 1}.</math>
| |
| − | | |
| − | ~UnknownMonkey, apex304, MRENTHUSIASM
| |
| − | | |
| − | ==Solution 2==
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| − | We label the octohedron going triangle by triangle until we reach the <math>?</math> triangle. The triangle to the left of the <math>Q</math> should be labeled with a <math>6</math>. Underneath triangle <math>6</math> is triangle <math>5</math>. The triangle to the right of triangle <math>5</math> is triangle <math>4</math> and further to the right is triangle <math>3</math>. Finally, the side of triangle <math>3</math> under triangle <math>Q</math> is <math>2</math>, so the triangle to the right of <math>Q</math> is <math>\boxed{\textbf{(A)}\ 1}</math>.
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| − | | |
| − | ~hdanger
| |
| − | | |
| − | ==Solution 3 (Fast and Cheap)==
| |
| − | Notice that the triangles labeled <math>2, 3, 4,</math> and <math>5</math> make the bottom half of the octahedron, as shown below:
| |
| − | <asy>
| |
| − | /*
| |
| − | Diagram by TheMathGuyd
| |
| − | Edited by MRENTHUSIASM
| |
| − | */
| |
| − | import graph;
| |
| − | // The Solid
| |
| − | // To save processing time, do not use three (dimensions)
| |
| − | // Project (roughly) to two
| |
| − | size(15cm);
| |
| − | pair Fr, Lf, Rt, Tp, Bt, Bk;
| |
| − | Lf=(0,0);
| |
| − | Rt=(12,1);
| |
| − | Fr=(7,-1);
| |
| − | Bk=(5,2);
| |
| − | Tp=(6,6.7);
| |
| − | Bt=(6,-5.2);
| |
| − | dot(Bt,linewidth(5));
| |
| − | draw(Lf--Fr--Rt);
| |
| − | draw(Lf--Tp--Rt);
| |
| − | draw(Lf--Bt--Rt);
| |
| − | draw(Tp--Fr--Bt);
| |
| − | draw(Lf--Bk--Rt,dashed);
| |
| − | draw(Tp--Bk--Bt,dashed);
| |
| − | label(rotate(-8.13010235)*slant(0.1)*"$Q$", (4.2,1.6));
| |
| − | label(rotate(21.8014095)*slant(-0.2)*"$?$", (8.5,2.05));
| |
| − | pair g = (-8,0); // Define Gap transform
| |
| − | real a = 8;
| |
| − | draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow
| |
| − | // Time for the NET
| |
| − | pair DA,DB,DC,CD,O;
| |
| − | DA = (4*sqrt(3),0);
| |
| − | DB = (2*sqrt(3),6);
| |
| − | DC = (DA+DB)/3;
| |
| − | CD = conj(DC);
| |
| − | O=(0,0);
| |
| − | transform trf=shift(3g+(0,3));
| |
| − | path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB);
| |
| − | dot(trf*(-DB),linewidth(5));
| |
| − | draw(trf*NET);
| |
| − | label("$7$",trf*DC);
| |
| − | label("$Q$",trf*DC+DA-DB);
| |
| − | label("$5$",trf*DC-DB);
| |
| − | label("$3$",trf*DC-DA-DB);
| |
| − | label("$6$",trf*CD);
| |
| − | label("$4$",trf*CD-DA);
| |
| − | label("$2$",trf*CD-DA-DB);
| |
| − | label("$1$",trf*CD-2DA);
| |
| − | </asy>
| |
| − | Therefore, <math>\textbf{(B)}, \textbf{(C)}, \textbf{(D)},</math> and <math>\textbf{(E)}</math> are clearly not the correct answer. Thus, the only choice left is <math>\boxed{\textbf{(A)}\ 1}</math>.
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| − | | |
| − | ~andy_lee
| |
| − | | |
| − | ==Video Solution by Math-X (Simple Visualization)==
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| − | https://youtu.be/Ku_c1YHnLt0?si=XilaQFDcnGR8ak7W&t=3364 ~Math-X
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| − | | |
| − | | |
| − | ==Video Solution (CREATIVE THINKING!!!)==
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| − | https://youtu.be/tcGcqm5RHiy
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| − | | |
| − | ~Education, the Study of Everything
| |
| − | | |
| − | ==Animated Video Solution==
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| − | https://youtu.be/ECqljkDeA5o | |
| − | | |
| − | ~Star League (https://starleague.us)
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| − | | |
| − | ==Video Solution by OmegaLearn (Using 3D Visualization)==
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| − | https://www.youtube.com/watch?v=nVSF2ujZkPE&ab_channel=SohilRathi
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| − | | |
| − | ==Video Solution by Magic Square==
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| − | https://youtu.be/-N46BeEKaCQ?t=3789
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| − | ==Video Solution by Interstigation==
| |
| − | https://youtu.be/DBqko2xATxs&t=2195
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| − | | |
| − | ==See Also==
| |
| − | {{AMC8 box|year=2023|num-b=16|num-a=18}}
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| − | {{MAA Notice}}
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