194Author(s): Marcus BakerSource: The American Mathematical Monthly, Vol. 10, No. 4 (Apr., 1903), p. 104Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2969225 .

Accessed: 14/05/2014 00:29

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org.

.

Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toThe American Mathematical Monthly.

http://www.jstor.org

This content downloaded from 194.29.185.23 on Wed, 14 May 2014 00:29:34 AMAll use subject to JSTOR Terms and Conditions

104

=3(R-h)2 (2R+h)/6R3 zm/n. Therefore, (h/R)3-3(h/R) +2=2m/n; or, (h1/R)3 -3(h/R)+2(n-m,)/n=-O. Applying the proper trigonometric formula, h/R=2V/(p/3)sinAO, where p=3, O=sin-13q/2p,/(3/p)=sin- 1(n-m)/n, q-= 2(n-m)/ni. Hence, h/R=2sin*[sin-'(n-m)/n].

194. Proposed by MARCUS BAKER. U. S. Geological Survey, Washington. D. C.

Glass paper weights, having the form of a regular tetrahedron, are to be packed for shipment, each in a paper box. Wanted to know the size and shape of the smallest box for the purpose. 1low much empty space in each box?

Solution by the PROPOSER.

Shape of box=cube.

Edge of box=s1/j, where s-edge of tetrahedron.

Empty space =sS 3/J.

Occupied space-71S3 1/j--volume of tetrahedron.

Total space=jsl1,/J volurne of box.

Of the twelve diagonals in the six faces of the box, the six edges of the tetrahedron coincide with one in each face.

CALCULUS.

160. Proposed by B. F. FINKEL. A.M., M.So., Professor of Mathematics and Physics, Drury College.Spring- field, Mo.

A dog at the vertex of a right conical hill pursues a fox at the foot of the hill. How far will the dog run to catch the fox, if the dog runs directly towards the fox at all times, and the fox is continually running around the hill at its foot, the velocity of the dog being 6 feet per second, the velocity of the fox being 5 feet per second, the hill being 100 feet high and 200 feet in diameter at the base?

Solution by G. B. M. ZERK, A. M., Ph.D., Professor of Chemistry and Physics, The Temple College, Philadel- pbia, Pa.

Let the origin be the vertex of the cone, 0 the center of the base of the cone, cr=the length of the dog's path, s=the length of the projection of the dog's path on the plane (x, y) or the base of the cone, rz-- radius vector of this projection, a=100 feet=altitude =radius of base, m-6 feet per second, n-5 feet per second, n/m__u, x'+y2=z' is the equation of the cone. Then uer-=zO, where e-')= Z COB, subtended by thie fox's path at the center 0.

da=(a/u)dO=kI(dx2 + dyo2+dz'2 )= I/(ds2+dz2)

-4[ra -(dr/do)2 +(dz/dO)2 ] da.

But r2 ,x +y5 =Z2. .-. dz=dr.

This content downloaded from 194.29.185.23 on Wed, 14 May 2014 00:29:34 AMAll use subject to JSTOR Terms and Conditions