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Question Number 189021 by mr W last updated on 10/Mar/23
How many non−similar triangles have integer angles in °?
$${How}\:{many}\:{non}−{similar}\:{triangles} \\ $$$${have}\:{integer}\:{angles}\:{in}\:°? \\ $$
Commented by nikif99 last updated on 11/Mar/23
Now I think there are 2700 solutions for ∡A, ∡B, ∡C integers degrees.
$${Now}\:{I}\:{think}\:{there}\:{are}\:\mathrm{2700}\:{solutions}\: \\ $$$${for}\:\measuredangle{A},\:\measuredangle{B},\:\measuredangle{C}\:{integers}\:{degrees}. \\ $$
Commented by mr W last updated on 11/Mar/23
that′s correct. can you share how you got this result?
$${that}'{s}\:{correct}.\:{can}\:{you}\:{share}\:{how} \\ $$$${you}\:{got}\:{this}\:{result}? \\ $$
Commented by nikif99 last updated on 11/Mar/23
∡A takes values 1 to 60 (to avoid circular repetitions), ∡B takes values from A to int(((180−A)/2)) and ∡C the rest till 180. ∡A=1 ∡B=1 to 89 89 cases ∡A=2 ∡B=2 to 89 88 cases ∡A=3 ∡B=3 to 88 86 cases (missing 87) ∡A=4 ∡B=4 to 88 85 cases ∡A=5 ∡B=5 to 87 83 cases (missing 84) ... ∡A=58 ∡B=58 to 61 4 cases ∡A=59 ∡B=59 to 60 2 cases (missing 3) ∡A=60 ∡B=60 to 60 1 case Σ_(n=1) ^(89) =((n(n+1))/2)=4005 less cases 87−84−81−...−6−3= 3Σ_(n=1) ^(29) =1305 ⇒real cases=4005−1305=2700
$$\measuredangle{A}\:{takes}\:{values}\:\mathrm{1}\:{to}\:\mathrm{60}\:\left({to}\:{avoid}\:\right. \\ $$$$\left.{circular}\:{repetitions}\right),\:\measuredangle{B}\:{takes}\: \\ $$$${values}\:{from}\:{A}\:{to}\:{int}\left(\frac{\mathrm{180}−{A}}{\mathrm{2}}\right)\:{and} \\ $$$$\measuredangle{C}\:{the}\:{rest}\:{till}\:\mathrm{180}. \\ $$$$\measuredangle{A}=\mathrm{1}\:\:\:\measuredangle{B}=\mathrm{1}\:{to}\:\mathrm{89}\:\:\mathrm{89}\:{cases} \\ $$$$\measuredangle{A}=\mathrm{2}\:\:\:\measuredangle{B}=\mathrm{2}\:{to}\:\mathrm{89}\:\:\mathrm{88}\:{cases} \\ $$$$\measuredangle{A}=\mathrm{3}\:\:\:\measuredangle{B}=\mathrm{3}\:{to}\:\mathrm{88}\:\:\mathrm{86}\:{cases}\:\left({missing}\:\mathrm{87}\right) \\ $$$$\measuredangle{A}=\mathrm{4}\:\:\:\measuredangle{B}=\mathrm{4}\:{to}\:\mathrm{88}\:\:\mathrm{85}\:{cases} \\ $$$$\measuredangle{A}=\mathrm{5}\:\:\:\measuredangle{B}=\mathrm{5}\:{to}\:\mathrm{87}\:\:\mathrm{83}\:{cases}\:\left({missing}\:\mathrm{84}\right) \\ $$$$… \\ $$$$\measuredangle{A}=\mathrm{58}\:\:\:\measuredangle{B}=\mathrm{58}\:{to}\:\mathrm{61}\:\:\mathrm{4}\:{cases} \\ $$$$\measuredangle{A}=\mathrm{59}\:\:\:\measuredangle{B}=\mathrm{59}\:{to}\:\mathrm{60}\:\:\mathrm{2}\:{cases}\:\left({missing}\:\mathrm{3}\right) \\ $$$$\measuredangle{A}=\mathrm{60}\:\:\:\measuredangle{B}=\mathrm{60}\:{to}\:\mathrm{60}\:\:\mathrm{1}\:{case} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{89}} {\sum}}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{4005} \\ $$$${less}\:{cases}\:\mathrm{87}−\mathrm{84}−\mathrm{81}−…−\mathrm{6}−\mathrm{3}= \\ $$$$\mathrm{3}\underset{{n}=\mathrm{1}} {\overset{\mathrm{29}} {\sum}}=\mathrm{1305}\:\Rightarrow{real}\:{cases}=\mathrm{4005}−\mathrm{1305}=\mathrm{2700} \\ $$
Commented by mr W last updated on 11/Mar/23
thanks alot for your nice solution!
$${thanks}\:{alot}\:{for}\:{your}\:{nice}\:{solution}! \\ $$
Answered by nikif99 last updated on 11/Mar/23
675 found using sin A+sin B+sin C=4cos (A/2) cos (B/2) cos (C/2) and triangle inequalities.
$$\mathrm{675}\:{found}\:{using}\: \\ $$$$\mathrm{sin}\:{A}+\mathrm{sin}\:{B}+\mathrm{sin}\:{C}=\mathrm{4cos}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{C}}{\mathrm{2}} \\ $$$${and}\:{triangle}\:{inequalities}. \\ $$
Commented by mr W last updated on 11/Mar/23
please explain a little more! the angles of the triangle should be integers, e.g. 1°+2°+177° is such a triangle.
$${please}\:{explain}\:{a}\:{little}\:{more}! \\ $$$${the}\:{angles}\:{of}\:{the}\:{triangle}\:{should}\:{be} \\ $$$${integers},\:{e}.{g}.\:\mathrm{1}°+\mathrm{2}°+\mathrm{177}°\:{is}\:{such}\:{a} \\ $$$${triangle}. \\ $$
Commented by nikif99 last updated on 11/Mar/23
please let me examine it a little more. it seems there are errors.
$${please}\:{let}\:{me}\:{examine}\:{it}\:{a}\:{little}\:{more}. \\ $$$${it}\:{seems}\:{there}\:{are}\:{errors}. \\ $$
Answered by mr W last updated on 11/Mar/23
an other method using generating function: α+β+γ=180° with 1°≤α≤β≤γ let α=1+p with p≥0 β=α+q=1+p+q with q≥0 γ=β+r=1+p+q+γ with r≥0 (1+p)+(1+p+q)+(1+p+q+r)=180 ⇒3+3p+2q+r=180 ...(i) with p,q,r≥0 number of solutions of (i) is the coef. of term x^(180) in the expansion of x^3 (1+x^3 +x^6 +...)(1+x^2 +x^4 +...)(1+x+x^2 +...) =(x^3 /((1−x^3 )(1−x^2 )(1−x))) which is 2700. therefore there are 2700 triangles with integer angles in degrees.
$${an}\:{other}\:{method}\:{using}\:{generating} \\ $$$${function}: \\ $$$$\alpha+\beta+\gamma=\mathrm{180}° \\ $$$${with}\:\mathrm{1}°\leqslant\alpha\leqslant\beta\leqslant\gamma \\ $$$${let} \\ $$$$\alpha=\mathrm{1}+{p}\:{with}\:{p}\geqslant\mathrm{0} \\ $$$$\beta=\alpha+{q}=\mathrm{1}+{p}+{q}\:{with}\:{q}\geqslant\mathrm{0} \\ $$$$\gamma=\beta+{r}=\mathrm{1}+{p}+{q}+\gamma\:{with}\:{r}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{1}+{p}\right)+\left(\mathrm{1}+{p}+{q}\right)+\left(\mathrm{1}+{p}+{q}+{r}\right)=\mathrm{180} \\ $$$$\Rightarrow\mathrm{3}+\mathrm{3}{p}+\mathrm{2}{q}+{r}=\mathrm{180}\:\:\:…\left({i}\right) \\ $$$${with}\:{p},{q},{r}\geqslant\mathrm{0} \\ $$$${number}\:{of}\:{solutions}\:{of}\:\left({i}\right)\:{is}\:{the} \\ $$$${coef}.\:{of}\:{term}\:{x}^{\mathrm{180}} \:{in}\:{the}\:{expansion}\:{of} \\ $$$${x}^{\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{3}} +{x}^{\mathrm{6}} +…\right)\left(\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +…\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right) \\ $$$$=\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}^{\mathrm{3}} \right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}\right)} \\ $$$${which}\:{is}\:\mathrm{2700}. \\ $$$${therefore}\:{there}\:{are}\:\mathrm{2700}\:{triangles} \\ $$$${with}\:{integer}\:{angles}\:{in}\:{degrees}. \\ $$
Commented by mr W last updated on 11/Mar/23

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