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Case VI.
Let the given ratio of EL to LI be inæqualitatis majoris, and
let the point ſought be required to lie beyond either extreme. Here we muſt
uſe the IIId Case of the IId Problem; and the Determination is that
UN (found in the ſame ratio to AI as IL is to IE) muſt not be leſs than
IE + EL + √4 IEL*.
let the point ſought be required to lie beyond either extreme. Here we muſt
uſe the IIId Case of the IId Problem; and the Determination is that
UN (found in the ſame ratio to AI as IL is to IE) muſt not be leſs than
IE + EL + √4 IEL*.
Case VII.
Let the ſituation of O be required the ſame as in the two laſt
Caſes, but let the given ratio be that of equality, which was there ſuppoſed
of inequality. Here the IId Problem will be of no uſe, and this Caſe
requires a particular conſtruction.
Caſes, but let the given ratio be that of equality, which was there ſuppoſed
of inequality. Here the IId Problem will be of no uſe, and this Caſe
requires a particular conſtruction.
Let then the three Points be A, I, E, and I the middle one;
and let it
be required to find a fourth O beyond E, ſuch that AO x OE may equal
OI2.
be required to find a fourth O beyond E, ſuch that AO x OE may equal
OI2.
Construction.
Upon AE diameter deſcribe a circle,, and let another YS
cut the former at right angles. Join SI, and continue it to meet the circum-
ference in R. From R draw a tangent to meet the given line in O, and I
ſay O is the point required.
cut the former at right angles. Join SI, and continue it to meet the circum-
ference in R. From R draw a tangent to meet the given line in O, and I
ſay O is the point required.
Demonstration.
Joining YR, the triangles SUI and SYR will be ſimi-
lar, and the angle UIS or RIO = SYR. But the angle IRO made by the
tangent and ſecant = SYR in the alternate ſegment. Therefore RIO = IRO,
and OR = OI. But by the property of the circle AOE = OR2. And
therefore AOE = OI2.
lar, and the angle UIS or RIO = SYR. But the angle IRO made by the
tangent and ſecant = SYR in the alternate ſegment. Therefore RIO = IRO,
and OR = OI. But by the property of the circle AOE = OR2. And
therefore AOE = OI2.
Q.
E.
D.
The Determination is that AI muſt be greater than IE.
Case VIII.
Whereas in the Iſt and IId Cases the given ratio was that of
inequality, let us now ſuppoſe it that of equality; and let the three points be
A, E, I, and E the middle one; and let a fourth O be ſought between E and
I, ſuch that AOE may equal OI2.
inequality, let us now ſuppoſe it that of equality; and let the three points be
A, E, I, and E the middle one; and let a fourth O be ſought between E and
I, ſuch that AOE may equal OI2.
The Construction and Demonstration of this Caſe is in every reſpect
the ſame as that of the preceeding, as will appear by comparing the figures.
the ſame as that of the preceeding, as will appear by comparing the figures.
PROBLEM IV.
To cut a given indefinite right line ſo in one point that, of the four ſeg-
ments intercepted between the ſame and four points given in the indefinite
line, the rectangle under any two aſſigned ones may be to the rectangle under
the two remaining ones in a given ratio.
ments intercepted between the ſame and four points given in the indefinite
line, the rectangle under any two aſſigned ones may be to the rectangle under
the two remaining ones in a given ratio.
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