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Bài giải: Toán thi vào đại học Nhật Bản

[ Saturday, 26 March, 2011 ]

Tokyo University’s entrance exam 2008

A regular octahedron is placed on a horizontal rest. Draw the plan of top-view for the regular octahedron.
Let G1 and G2 be the barycenters of the two faces of the regular octahedron parallel to each other. Find the volume of the solid by revolving the regular tetrahedron about the line G1G2 as the axis of rotation.


First, a regular octahedron is a Platonic solid composed of eight equilateral triangles, four of which meet at each vertex. Let the regular octahedron be ABCDTS where ABCD is the square in the middle, T and S are the vertices at top and bottom, respectively.
By definition, these triangles are equilateral TAB, TBC, TCD, TAD, SAB, SBC, SCD and TAD.

Let [Φ] denote the plane containing shape Φ, M and N be the midpoints of AB and CD, respectively, G1 and G2 be the barycenters located on the parallel triangles TAB and SCD, respectively, P and Q the midpoints of G1G2 and TD, repectively. Now let a be the side length of the equilateral triangles, = ∠MTN, = ∠TNG1 and =∠TMN = ∠TNM = ∠MNS = ∠NMS = 90° –2 .
Since M and N are the midpoints, we have TM = TN = ½a3, and with G1 and G2 being the barycenters, we get TG1 = 23TM = SG2 = a/3. Now let’s find cos. Applying the law of cosines, we obtain
MN² =TM² + TN² – 2TM×TN×cos, or a² = 32a²(1 – cos), or
cos = 13. We need to verify that G1G2 is perpendicular to both [TAB] and [SCD]. Armed with the value of cos, we can get the length of segment G1N which is now
G1N² = TN² + TG1² – 2TN×TG1×cos= 34a², or G1N = TN and thus ∠TG1N = ∠MTN = .

We also have ∠G1NG2= ∠TNS –  = 2–  = 180° – –  = . Combining with the fact that the points T, G1, M, S, G2 and N are coplanar and ∠TG1N = , TM is parallel to SN. We now find the length G1G2.
G1G2² = G1N² + G2N² – 2G1N×G2N×cos = 23a², or G1G2 = 26 a.
It follows that G1N² = G1G2² + G2N² = 34a², or G1G2 ⊥SN. Because TM || SN, we also have G1G2 ⊥TM.

Now since [TG1MSG2N] ⊥CD, we get G1D = G1C = G1N² + DN² = a = CD. The triangle CG1D is congruent with other triangles of this regular octahedron, and since G2 is the barycenter of an equilateral triangle, G2D = SG2 = a/3. We now have G1D² = a² = G1G2² + G2D², or G1G2 ⊥G2D.

Combining with G1G2⊥SN, we conclude that G1G2 is perpendicular to both parallel planes [TAB] and [SCD], and also combining with the fact that P is the midpoint of G1G2, TP = BP = AP = PC = PD = PS.

Using G1G2 as the axis of rotation, we now need to find the farthest points away from G1G2 of this regular octahedron, and they are T and S which are equidistant from G1G2. Meanwhile, Q and the midpoint of SB are the shortest and equidistant from G1G2. When revolving the octahedron forms a volume that is symmetrical with respect to the revolving line PQ, and the total volume equals twice the volume obtained by revolving TM and PQ as seen on the graph on the next page.

Now let’s find PQ. As previously discovered, TP = PD and
TP = TG1² + PG1² = TG1² + ¼G1G2² = a/2, TQ = ½TD = a/2. Therefore, PQ = TP² – TQ² = a/2.

The volume of the regular octahedron when revolving is now
V = 2×13R²(d + a6) – r²d ] where R = G1T, r = PQ, d is the distance from the intersection of the extensions of TQ and G1G2 to P. Because of the similar triangles, we have

dPQ = PG1 TG1 – PQ , or d = a(2 + 3)2.

Finally, V = [1/18sqrt(6)]pi*a³][7 + 2sqrt(3)].

Note: This website does not translate formulas from Microsoft Word correctly and many results in the above equations are not correct. The author of the solution is not able to correct them. However, the overall strategy and method to solve are correct.

We’re lucky we’re not Japanese. It’s not only about earthquakes and tsunamis, but it’s also these types of exams required to enter colleges. Many Japanese students are known to commit suicide when not admitted to college.

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