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A more elementary but less fortunate method consists in using (1) and the corresponding relation Q-2hb2÷(b+h)

..........

Now from (1), h(2a2-P2)=P2a. But h2=a+b2. Hence

(8).

b2=4a4 (P2 — á2) ÷ (2a2 — P2) 2.

(9).

Eliminating between (2) and (8), we get

Pa2 (2b2-Q2)2=Q+b2 (2a2 - P2)2.

In this we substitute the value of b2 from (9) and obtain an equation of the sixth degree for a2. Set a=2a2-P2. Then

(P++Q1)«®+2P1 (P2 +Q2)a +P° (2Q2 - P2)a4-2P8 (2P2+Q2) a3

-P10 (20+P2)a2 +2P14a2 + P100......

(10).

We may take PQ. Then by Descartes' Rule of Signs, there are two or no positive roots. There are two positive roots, so that (10) does not uniquely deter

mine the leg a.

218. Proposed by O. W. ANTHONY, DeWitt Clinton High School, New York City.

From a given triangle cut off an area equivalent to a given square by a line passing through a given point without the triangle.

IV. Solution by G. B. M. ZERR, A. M., Ph. D., Parsons, W. Va.

Let ABC be the given triangle, MN a side of the given square, P the given

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point. Through P draw PE parallel to AC-
meeting AB in E. Perpendicular to AB
draw AD-MN, lay off AF-MN. Join
DE and draw FO parallel to DE, cutting
AC in G'. Draw OGH parallel to AB.
At G erect GJ PE perpendicular to AC
and draw JK-PH. Draw PLIK.

From similar triangles AED and
AFO, we have AE: AD-AF(=AD): AO.
.. AD AEX AO-area AEHG; JK2-
GK or PHJ-PEL-GIK
MN.

[graphic]

JG2
..ALM

V. Solution by J. SCHEFFER, A. M.. Hagerstown, Md.

=

LEHI.

Let ABC be the triangle and P the given point without. Draw PE parallel to AB, cutting AC in D. Make parallelogram DEFA = given square. On F erect the perpendicular FG-PD, and make GQ-PE. Connect P with Q, then will PQ be the required line.

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31; in Beшan and Smith's translatión, p. 34.

II. Solution reported by the PROPOSER.

The elegant solution by Lill (reported without proof by d'Ocagne at the Second International Congress of Mathematicians, Paris, 1900) is so simple that

Y Q

B

A

OM

NH 2

the Proposer has used it in his courses in elementary algebra. For the graphic solution of x2+px+q=0, choose two perpendicular lines Ox and Oy, lay off unit length OA on Oy, length OH on Ox containing -p units (to right or left of O, according asp is or ), length HB on parallel to Oy containing q units. If the circle on AB as diameter cuts Ox at M and N, then OM and ON, on the same scale, are the required roots. In proof, let Q be the second point of intersection of the circle Oy, then OQ=HB, since OHBQ is a rectangle; OM

-NH by eqnality of triangles OQM and HBN. Hence OM. ON=OA.OQ=q, OM+ON OH=-p.

III. Solution by B. F. FINKEL, A. M., M. Sc., 204 St. Marks Square, Philadelphia, Pa.

Let ax + bx+c=0, be the general quadratic. On the line AD, lay off AB-2 units, and BD=c/2a. On AD as a diameter describe the circle AED. At B erect the perpendicular BE. With E as a center and a radius equal to b/2a, describe an are intersecting AD, or, AD produced, in C. Then with C as a center and a radius equal to CB describe the circle FBG intersecting EC in G and F in the order.

[graphic]

E, G, C, F. Then EG is equal to the value of one root of the quadratic and EF is equal to the other. For

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This construction only gives the absolute values of the real roots.
responding algebraic values must be assigned.

If b2/4a2=c/a, then CE-BE, which equals x.
If b2/4a2 <c/a, the construction is impossible.

IV. Solution by A. H. HOLMES, Brunswick, Maine.

The corres

Describe a circle of radius a, and from a point on the circumference A draw tangent AB/b. Then from B draw through the center of the circle O the line BD cutting the circumference in C and D. By the principles of plane geometry the lines CB and BD will represent the unknown quantity in the quadratic equation x2+2ax=b.*

V. Solution by L. LELAND LOCKE, Brooklyn, N. Y.

We have four cases to consider. These may be reduced to two.

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[graphic]

(3),

(4).

Let q=r.s, r and s being any suitable factors of q.

Case I. Equations 1, 2. Draw a circumference 1, and in it a chord AB=p, using any convenient unit of measure. Tangent to this chord and concentric with circle 1 draw circle 2. In circle 1 place a chord MNP such that MP=r and NP-s., Through P draw chord CD tangent to circle 2. PC and PD are the roots of the quadratics. Their values may be found, using the same unit of measure as before.

Proof. AB-CD=p, CP=x,, PD=x-p-x; CP.PD-NP.PM=s.r, x(p-x)=rs,

x-px1+rs-07

2 1

Similarly, x-px+rs=0 equation 1.

*There are four cases, since a and b may each be positive or negative. The above solution suggests at once the following: Construct a circle of radius a; at the distance /b from the diameter AB draw a parallel chord which intersects the circle (if at all) in A'B'. Draw B'H perpendicular to AB; it is now evident that AH, HB are roots of the quadratic (2a-x)x-b; i. e. of x2-2ax+b=0.

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In order to solve x2-2ax-b we observe that by changing the signs of both roots the equation 2+2ax =b is obtained, and Mr. Holmes' solution applies.

Similarly, in order to solve x2+2ax+b=0 we observe that its roots are the negatives of the roots of x2-2ax+b=0, and the construction at the beginning of this foot-note applies. ED.

Equation 2, x2-px2+q=0, may be solved in a similar manner by changing the sign of px and proceeding as above. The roots being the same in numerical value but opposite in sign.

Case II. Equations 3, 4. The figure for Case II differs from that of Case I only in that P is outside of the circle 1; the points CPD being now in the order POD. PC, PD-x=x+p, PN.PM-PC.PD; r.s=x(x1+p), x2+px, -rs =0. Similarly, x2+px, -rs-0.

2

1

2 1

1

Equation 4. is solved by changing sign of pr and proceeding as with equation 3, since the roots are the opposite of the roots of equation (3).

In Case I if the roots are imaginary, the point P falls within circumfer

ence 2, and the graphic method fails.

VI. Solution by G. B. M. ZERR. A. M., Ph. D., Parsons. W. Va.

Upon AB describe a semi-circle. Let C be the mid-point of the semi-circular arc. Draw AC, BC. Let G be a point on the line AC; on GC describe a circle center D. Through D draw BFDE. Then taking BE, BF positive; EB, FB negative, these lines are the roots of a quadratic having 2CD for the coefficient of the first power of the unknown quantity, and 1/(2AB) for the absolute term, the coefficient of the second power of the unknown quantity being taken unity..

[graphic]

Let AB=1/2c, GO-2D0=b.
BE=BD+ DE=BD+DC=√(BC2+DC2)+D0=3[b+√(b2+4c)].
BF-BD-DF-BD-DO--[b-v (b2+4c)]. Taking xbx-c.

2

then for x2+bx-c, EB=-[b+√(b2+4c)], BF=-1[b—√(b2+4c)].
For x2 bx-c, BE [b+1/(b2+4c)], FB={[b−1/(b2+4c)].
If c be negative, the results still hold.

Also solved by G. W. Greenwood, B. A. (Oxon), Professor of Mathematics, and Astronomy in McKendree College, Lebanon, Ill., by use of circle and hyperbola.

219A. Proposed by H. F. MacNEISH, A.B., Assistant in Mathematics, University High School, Chicago, Ill. Draw a line through a given point which shall divide a given quadrilateral into two equivalent parts; (1) when the point lies in a side of the quadrilateral, (2) when the point is without, (3) within the quadrilateral..

F.

Solution by G. B. M. ZERR. A. M., Ph. D.. Parsons. W. Va.

Let ABCD be the given quadrilateral. Produce DA, CB till they meet in Join AC and draw BS parallel to AC, join CS; then triangle SCD=quadriABCD. Disect SD in H, HF in G, and join CH, CG.

(1). Let P be the point in the side BC. Join PH and draw CL parallel

to PH, join PL. Triangle PCL-triangle HCL.

..POL+LCD PODL=HCL+LCD-HCD-ABCD.
(2). Let R be the point without the quadrilateral.

Draw RK parallel to

FD meeting FC in K.

Join KG and draw CI parallel to KG, join KI. Then

ICG-ICK and FCI=FCI..

..FCG=FKI. Draw IJ parallel to FC, then parallelogram FKJI=FCH. At the point I draw IZ perpendicular to FD and equal to RK, draw ZL-RJ.

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:. FPL-FENM-FCH... FPL-FBA=FCH-FCS-SCH.
..ABPL-SCH-ABCD::

Also solved by G. W. Greenwood, B. A. (Oxon).

220. Proposed by G. B.M. ZERR, A. M., Ph. D., Parsons, West Va.

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Two triangles are circumscribed to a given triangle ABC, having their sides perpendicular to the sides of the given triangle. Prove that the two triangles are equal, and find the area of these triangles.

I. Solution by G. W. GREENWOOD, B. A. (Oxon), Professor of Mathematics and Astronomy, McKendree College, Lebanon, Ill.

The perpendiculars to a side of the given triangle at its extremities, which are corresponding sides of the circumscribed triangles, are symmetrical with respect to any point on the perpendicular bisector of that side. Hence the two triangles are symmetric with respect to the circumcenter of the given triangle and are therefore equal. The area of either is equal to

(c2 cotB+a2cotC+b2 cotA)+S,

where S is the area of the given triangle.

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