L'ho risolto con la geometria analitica...la retta y=mx+10 è tangente alla circonferenza (x-5)^2+y^2=25..imponendo Δ=0,risulta m=-3/4..calcolo adesso il punto M,come intersezione tra y=(-3/4)x+10 e la circonferenza x^2+y^2=100=>M=(48/5,14/5).x=√((48/5)^2+(10-14/5)^2)=12
Draw a square AOBC. Let R be the intersection of straight lines BC and AN. AO:OP=PB:BR => BR=5/2 AO:BR=4:1 , BN=y => (10+y):y=4:1 => y=10/3 , PQ:BR=2:1 => QN=2*BN=20/3
Extremely easy:
tanα = r/R= 5/10= 1/2 --> α= 26,565°
x = 2R cos2α = 20*3/5= 12 cm
Agree completely. But better to find tan2a = (1/2+1/2)/(1- 1/4) = 4/3, then cos 2a =3/5. More elegant.
Let I be the center of the semicircle. We put
At 3:25, construct AP. ΔAOP and ΔAQP are congruent by side-side-side. By congruency,
L'ho risolto con la geometria analitica...la retta y=mx+10 è tangente alla circonferenza (x-5)^2+y^2=25..imponendo Δ=0,risulta m=-3/4..calcolo adesso il punto M,come intersezione tra y=(-3/4)x+10 e la circonferenza x^2+y^2=100=>M=(48/5,14/5).x=√((48/5)^2+(10-14/5)^2)=12
10*cos(
Draw a square AOBC.
Let R be the intersection of straight lines BC and AN.
AO:OP=PB:BR => BR=5/2
AO:BR=4:1 , BN=y => (10+y):y=4:1 => y=10/3 , PQ:BR=2:1 => QN=2*BN=20/3
PAO = QAP = δ → AM = AQ + MQ = 10 + (x - 10); AR = AO + RO = 20; sin(BOR) = 1
sin(δ) = √5/5 → cos(δ) = 2√5/5 → sin(2δ) = 2sin(δ)cos(δ) = 4/5 →
cos(2δ) = 3/5 = x/20 → x = 12 → x - 10 = 2 = MQ; PM = √29
(10)^2=100/90=1.10.1.2^5 1.2^1 2^1 (AMPO ➖ 2AMPO+1).
12
I am getting x as 12.5 lol
So close yet so far
Me too,but MB isn't tangent, perpendicular to BO.