224 L. EULERI OPERA POSTHUMA. Arithmetka. 



60. > 



Probleua. Invenire duo triangula rectangula in numeris, quoruin areae 



Az=pq{^ — ??) et B=:rs {rr — ss) 



A a 



datam inter se teneant rationem scil. a-.h, ita ut sit — -=— . 



B 



Hoc problema methodo direcU frustra tractatur, unde ad solutiones particulares confugere necesse est 

 cujusmodi sunt sequentes : 



I. Sumatur r=f et s=p — g, erit rH-s = 2p — q et r — s = q, unde fit 



• '■B==y{V--^i)[2'p—q)q 

 hincque -^ = f-^*- = 4' ^^^^ hp-\rhq = 2aj^ — aq, ideoque — = ^^_^ , unde haec solutio aaficitur 



|)= a ~\-h r= a -\~h '■•'' '"'' 



qz=2a — h s = 2h — a. 



II. Sit r = 2p et s = p-¥-q, erit rH-s = 3p-+-g et r — s=p — q, hino -fi = 2p(p-+-5') (3|)-4-7) (p— ?), ergo 



= =— , sicque erit 6^ = 6ap-i-2ag^, indeque — = 



6-2o 



quocirca capiatar p = b — 2a r = 2h — ka 



q = Ga ' ■ 8 = h -i-4-a. 



III. Sit r = 2p et s=p — g, erit rH-s = 3p — g, et r — s=p-t-?, hinc 



■F=6^:r2^=6-' *'"*1"^ hq = Qap-2aq, ergo ^ = -6^' 

 quocirca capiatur pro ista solutione p = 6-f-2a r = 26-f-4a 



q = Qa s = h — ia. 



IV. Sit r=p-i-j et s=p, eritr-t-s = %)-!-} et r-—s = q. hincque 



T-=2^=T' ""^^ ^^"*^^"^^^ '*P-*^ = 2«-P-^^^' 7=6^ 

 ergo capiantur p = a-f- 6 r = 26 — a 



q = h — 2a s = a-4-6. 



V. Sit r=p-4-gr et s = q, erit rHh-s=p-i-22 et r-^s^p, unde fit 



A p — q »._.,, ^ ,, p 2a-i-6 



-— =^^- — ^=— , mde hp — hq = ap-\-2aq, hmc —=7 — — , 



quocirca capere debemu^ p = 2a-i-6 r = a-f-26 



i; ' Y. ' q= h — a s=^h — a. 



VI. Sit r=p-i-5f «t « = 2^, erit r-i-s=p-f-35 et r-^s^p — q, unde fit 



= ^, unde ob fcp = 2ap -1-605', erit — =r — g-. 



,-,. B 2i>-*^6? &' i i- -rf» j 6 — 2o 



ideoque hic capere oportet p = 6a r=^\a-\-b 



: g = i~2a s = 2b—h-a. 



VII. Sit r=p — g et s = q, erit r-i-s = p et r — s = p — 2g et erit 



i.»,Sk — ^^* ,i.i. v.ii- — i: — ^::^ ideoque hp-\-hq = ap — 2aq, hmc — = _, , 



itaque ut capiatur necesse est p = 6-*-2a r = 26-f-a 



q = a — h $= a — b. 



