ment. Similarly, angle ABA' is obtained. The supplement for dihedral angle OA can be obtained by a like construction, or more quickly as follows: Erect a perpendicular to OA, at A, to cut OC at P; and a perpendicular to OA at A to cut OB at Q. Figure 5 Make triangle cut out the sector A'OA', with a lug at A'. Draw lines OX perpendicular to OC', OX' perpendicular to O B', and so on as indicated by Fig. 5 following the notation of Fig. 2. Crease along OB' and OC' and fasten together at A'A'1. Now the two parts can be assembled and the model completed. Should the class contain a boy with the requisite mechanical skill, this model can be used as a pattern for a still better one made of wire. Figure 6 The angles as given here have been chosen so that the face angles on the second trihedral angle are each obtuse. When one of them is acute, certain difficulties arise in that a face of the first part intersects a face of the second part before it reaches the face to which it is normal. Again, when the initial angles are such that F falls, not in angle BOC but in its vertical angle, certain modifications of procedure are necessary. These diffiiculties can be surmounted without much trouble; to discuss them here would complicate a first presentation too much. PROBLEM DEPARTMENT. Conducted by J. O. Hassler. Crane Technical High School and Junior College, Chicago. This department aims to provide problems of varying degrees of difficulty which will interest anyone engaged in the study of mathematics. Besides those that are interesting per se, some are practical, some are useful to teachers in class work, and there are occasionally some whose solutions introduce modern mathematical theories and, we hope, encourage further investigation in these directions. We desire also to help those who have problems they cannot solve. Such problems should be so indicated when sent to the Editor, and they will receive immediate attention. Remember that it takes several months for a problem to go through this department to a published solution. Each All readers are invited to propose problems and solve problems here proposed. Problems and solutions will be credited to their authors. solution, or proposed problem, sent to the Editor should have the author's name introducing the problem or solution as on the following pages. In selecting problems for solution we consider accuracy, completeness, and brevity as essential. The Editor of this department desires to serve its readers by making it interesting and helpful to them. If you have any suggestion to make, mail it to the Editor. Address all communications to J. O. Hassler, 2337 W. 108th Place, Chicago. SOLUTION OF PROBLEMS. 571. Proposed by Harold M. Lufkin, St. Andrew's School, St. Andrew's, Tenn. No communications concerning this problem were received. The problem as it is stated has no solution. If cleared of fractions and radicals the value x 1/7 is found, which does not satisfy the equation. If the second radical is replaced by - √2x+1, 1/7 is a root.-Editor. = Geometry. 572. Proposed by Takeshi Omachi, Sendai, Japan. Catalan's Theorem: In a quadrilateral ABCD, if the sides AB and CD are equal to each other, the straight line MN passing through the middle points M and N of the sides BC and AD will equally incline to the sides AB and CD. I. Solution by Philomathe, Montreal, Canada. = Draw NE equal and parallel to AB, also NF equal and parallel to CD. Join BE, EF, FC; EF passes through M and EM MF (BEFC a parallelogram). Hence, in the isosceles triangle ENF, the line MN bisects the vertical angle N, that is MN is equally inclined to AB and CD. II. Solution by A. MacLeod, Aberdeen, Scotland. = DS . KQ = HR. Draw AP, BQ, CR, DS perpendicular to MN, AK perpendicular to BQ, and DH perpendicular to CR. From triangles APN, DSN, AP From triangles BQM, CRM, BQ CR .BK == CH. From triangles BAK, CDH, ZBAK nations of AB and CD to MN. = <CDH, and these are the incli Also solved by the PROPOSER, L. E. A. LING and R. M. MATTHEWS. 573. Proposed by Clifford N. Mills, Brookings, S. Dak. From one of the angles of a rectangle a perpendicular is drawn to its diagonal, and from the point of their intersection straight lines are drawn perpendicular to the sides which contain the opposite angle. Show that if a and b be the lengths of the perpendiculars last drawn, and c the diagonal of the rectangle, al+b} = cl. I. Solution by Philomathe, Montreal, Canada. Let m, n be the sides of the rectangle respectively parallel to b and a. Well known relations in the triangles give at once the equations, m2+n2 = c2, bc/ m2 = m/c, ac/n2 = n/c. Eliminate m and n from the system: II. Solution by A. MacLeod, Aberdeen, Scotland. Given: Rectangle ABCD, AO1BD, ОE1BC, OF 1CD. To prove if OE = a, OF = b, BD = c, then al+b} = ci. Let ZABD 0. Then DAO = 0. = III. ..a = ccos3 0. :.b = csin30. Solution by L. E. A. Ling, La Grange, Ill. Given rectangle ABCD, DE LAC, EM = b and 1BC; EH = a and LAB; HE intersecting CD in L, EL = k; ME intersecting AD in N, NE = g; AC = c. Also solved by R. M. MATHEWS, who appends the following remark: "When a line of constant length moves with its ends on two rectangular axes it envelopes a curve whose equation is x}+y} = c}. This problem furnishes a construction for the point of contact of the line with its envelope." 574. Proposed by N. P. Pandya, Amreli, Kathiawar, India. Construct a triangle ABC, having given the distances of its incentre, circumcentre, and orthocentre. Solution by the Editor. (No solutions received.) Let AI, AH, AO be given distances from vertex A of triangle ABC to incenter I, orthocentre H, and circum centre O, respectively. Lemma I. The distance AH is twice the distance from O to the side BC. Lemma II. If R and r be the radii of the circumcircle and incircle, respectively, and d(IO) the distance between the centers, d2 = R2-2rR. Proofs of these lemmas may be found in Vol. XV, pp. 258, 259 (March, 1915), SCHOOL SCIENCE AND MATHEMATICS, if not elsewhere. Choose a point O as center and draw circum circle with radius OA. With radius AH/2 and center O describe circle which (by Lemma I) will be tangent to side BC. This lies within circum circle, for the orthocenter of any triangle ABC is the circumcenter of a triangle A'B'C' formed by drawing lines through A, B, C parallel respectively to BC, CA, AB. Then, AH<R' and R' = 2R, .AĤ/2<R. (R' circum radius of A'B'C'.) Draw chord of circumcircle tangent to inner circle and choose this as side BC of the desired triangle. Angle A is thus determined. Construct an angle equal to ZA, bisect it, choose length AI on bisector, and drop perpendicular from I to one side determining r Construct d, a mean proportional between R and R-2r. (Lemma II.) The intersection of a circle with radius d and center O with a line parallel to BC at the distance r from BC is the center I of the incircle. The intersection of a circle with center I and radius IA with the circum circle will give the third vertex A of the triangle. Proofs not contained in the statements are obvious. 575. Selected. Prove this theorem from Wentworth-Smith's Solid Geometry (Exercíses on Book VI) by means of earlier theorems. If the face angle AVB of the trihedral angle V-ABC is bisected by the line VD, the angle CVD is less than, equal to, or greater than half the sum of the angles AVC and BVC according as CVD is less than, equal to, or greater than a right angle. Solution by Philomathe. = In the plane CVD make angle DVC' equal to DVC. Trihedral angles V-ADC and V-BDC' are equal, hence angles BVC' and AVC are equal. Now, if angle CVD <90°, the trihedral angle V-BCC' gives angle CVC' <BVC'+BVC, or CVC' <AVC+BVC, that is CVD<Ï/2(AVC+BVC). Next, if angle CVD 90°, CVC' is a straight line, then BVC'+BVC 180°, or AVC+BVC+ 180°; hence CVD · 1/2(AVC+BVC). Finally, if angle CVD >90°, then CVC'>180°. through V, AV to VA', BV to VB', and DV to VD', CVD'<90°; hence CVC'<CVA'+CVB' in trihedral angle V-A'B'C..360°-CVC' > (180° - CVA')+(180° −CVB'), or CVC'(reflex) >CVA+CVB. = = Producing Also solved by L. E. A. LING. R. M. Mathews calls our attention to the fact that this was solved as No. 404 in January, 1915. The two solutions published then differ from the one above.-Editor. Explanatory Note-Problem 567. Contributed by Philomathe, Montreal, Can. 567. Proposed by N. P. Pandya, Amreli, Kathiawar, India. Circumscribe a triangle about a given circle, the ratio of the angle bisectors being known. In 1857-58, Catalan proposed the following problem: "To construct a triangle knowing its three bisectors." Nobody solved it. However, by analysis, Professor Barbarin proved that the solution depends upon an equation of the 12th degree! See Mathesis, 1896, pp. 143, 154; also Intermediaire des Mathematiciens, 1904, p. 229, Question 2771. Now, No. 567 is correlative to Catalan's problem, and the solution of the former presupposes that of the latter. Therefore, I do not think the problem geometrically possible. 569. Proposed by N. P. Pandya. The vertex A of a triangle is the center of a given circle. P and Q are points of intersection of AB, AC, respectively, with the circle. The tangents at P and Q divide the base in the ratios kland m : n, respectively. Construct the triangle ABC. No solutions received. The Editor submits the following: For the problem to have a unique solution, P and Q must be given. Assume the triangle constructed. Let M and N be points of intersection with BC of tangents at P and Q, respectively, and let O be the point of intersection of PM and PN. BM: MC = kl, and BN: NC m: n; PBM and NQC are right triangles; POQ = 180° - ZA. We begin by constructing a figure A'P'B'M'O'N'C'Q'A' similar to the one desired. = On a given line B'C' determine points M' and N' so that B'M' : M'C' k1 and B'N' N'C' m : n. Describe semicircles (on same side of M'N') on B'M' and N'C' as diameters. If M' lies between B' and N' the tangents at P' and Q' intersect M'N' before intersecting each other. In this case draw an arc on M'N' (on side of M'N' opposite the semicircles) in which would be inscribed the angle 180°-ZA. P' will lie on semicircle B'M', Q' on N'C', and O' on arc M'N'. Our problem, then, is to determine two equal lines O'P' and O'Q' so that M' lies on O'P' and N' on O'Q'. Assuming this done, according to the conditions of the problem we have the following relations: Choosing, for brevity, the notation a ZA, given); p O'N', T = c+t. t/2r = = 8/c,. = (By similar triangles) = = = ZP'M'B', 8 = 4Q'N'C' B'M'/2, r = N'C'/2, c = Q'N', Y = projection of t on M'N'; then M'N', ¿ = T/sinẞt/sinad/sinA, (sinA = sin[180° (a+8)] We use the last five equations to express y, 7 and c in terms of t. Since from (4), Tsina - tsinẞ 0, (7) if we multiply (7) by sina, (5) by cosa, add, reduce by addition formula for cosines, substitute values in (6) and transpose, we have ~ = 2pcosa = 2p√d2 - t'sin3A 2p/dvd2-tsin A Substituting now the values of y and from (8) and (9), enerá γ c = (2prd-Tyr)/pt. c = 2r/d(tsin A-cosA√d-tsin3A). (8) (9) (10) Substituting the values of (8), (9), (10) in (1), we obtain, after some reduction, From this form t may be constructed although the work involves thir teen separate constructions by compass and ruler such as fourth proportional, sum of squares, finding a linear ratio equal to a ratio of squares, and making ratios to replace sinA and cosA. With t as a radius and N' as center describe are intersecting arc M'N' at O'. O'P' and O'Q' will be the equal tangents to the circle A' at the points P' and Q', respectively, and will determine the circle. If R' be |