Different Ways of Solving a Geometric Task
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The paper discusses multiple geometric solutions to a specific task involving angles in an equal-angle triangle. Five distinct solutions are detailed, including geometric, trigonometric, and analytical methods using complex numbers. Additionally, the authors emphasize the crucial role of mathematics in education and society, advocating for a positive perception of the subject.
![International Journal of Smart Computing and Information Technology
2020, Vol. 1, No. 1, pp. 1–3
Copyright c
2020 by BOHR Publishers
www.bohrpub.com
Different Ways of Solving a Geometric Task
Dragan Obradovic1
and Lakshmi Narayan Mishra2∗
1
Elementary school “Jovan Cvijic”, Kostolac-Pozarevac, Serbia
2
Department of Mathematics, School of Advanced Sciences, Vellore Institute of Technology (VIT) University,
Vellore, Tamil Nadu, India
*Corresponding author. E-mail: lakshminarayanmishra04@gmail.com
Abstract. We present the solution in several ways, i.e., one solution from geometry, four solutions are clearly geometric, i.e.,
one is trigonometric and three are analytical (two on complex level and one in Deck level).
Keywords: Geometry, triangle, circle, equal-angle triangle, trigonometry, task, sinusoid theorem, coordinates, complex num-
bers, trigonometric identities
Task and solutions
Here are some solutions to the following task:
In an equal-angle triangle ABC with the base BC the angle at the
top A is equal to 80◦
. In the interior of the ABC triangle, the point
M is such that the angle ]MBC = 30◦
and angle ]MCB = 10◦
.
Find the angle of AMC.
Solution 1.
Let D be the height of top A on the base of the BC of triangle
ABC and let the point N be an intersection of AD and BM. Then
there is the angle ]CMN external to the triangle BMC is the same
]CMN = ]MBC + ]MCB = 40◦
. It is also ]NAC = 40◦
because it
is half the corner at the top of the equal-triangular triangle ABC.
Since N ∈ AD triangle BCN is equal, it is ]NCB = ]NBC = 30◦
.
Now,
]NCM = ]NCB − ]MCB = 20
We know that:
]ACN = ]ACB − ]NCB = 20
Now, we have
]MNC = 180◦
− ]NMC − ]NCM = 120◦
i
]ANC = 180◦
− ]NAC − ]NCA = 120
So, the MNC triangle is matched with the ANC triangle based on
the following conditions:
]ANC = ]MNC = 120◦
,
]ACN = ]MCN = 20◦
And NC = NC. From this coincidence, we conclude that AC =
MC is a triangle ACM equal angle with angle at the top ]ACM =
40◦
. Its base is ]AMC = ]MAC = 70◦
. Thus, we now complete
solution 1.
Solution 2.
Let O be the center of the circle k described around the triangle
BMC. As ]MOC = 2]MBC = 60◦
since they are angles ]MOC
and ]MBC, respectively central and peripheral over with the MC
circle k. Since MO = OC it follows that the triangle MOC is equi-
lateral. It is also ]BOM central when it is ]BCM peripheral over
with the tendency of the BM circle k, it is ]BOM = 2]BCM =
20◦
.
1](https://image.slidesharecdn.com/article1ijscit-210127093107/85/Different-Ways-of-Solving-a-Geometric-Task-1-320.jpg)
![2 Dragan Obradovic and Lakshmi Narayan Mishra
We now have ]BOC = 80◦
, so since there are basics equal
triangles ABC and OBC are equal (that is along BC), as their
corners are at the top equal (i.e., ]BAC = ]BOC = 80◦
) we hence
have ABC and OBC matched, so is AC = OC. From the fact that
the MOC triangle is equilibrium, we have MC = OC, hence, AC
= MC. Thus, we have the AMC triangle with an angle at the top
]MCA = 40◦
. Hence, ]AMC = ]MAC = 70◦
. Thus, solution 2 is
now complete.
Solution 2 can also be given in the following way, so in the
first one the plan comes with the properties of symmetrical fig-
ures.
Solution 3.
Let point O be symmetric with point A in relation to BC law.
Then it will be AC = OC and AB = OB, and since AB = AC we
have AB = AC = OC = OB. Wehave a circle with the center O and
radius OB. Since OB = OC we have C∈k. Let X be an arbitrary
point of the circle such that the points X and M are made from
different sides by BC.
Now,
1
2
]BXC = ]BOC = 40◦
.
Since ]BMC = 180◦
− ]MBC − ]MCB = 140◦
, we have ]BMC +
]BCC = 180◦
, so the BXCM four-blade is tetitive. Now, the
points B, X, and C already on the circle k follow that M be-
longs to k, so MO = OC. Now, ]MOC = 2]MBC = 60◦
, so
the MOC triangle is equilateral, thus it follows that MC = OC.
Since we already have AC = OC, it follows that MC = AC, so the
MAC triangle is equal, so as in the previous version we have that
]AMC = ]MAC = 70◦
.
Solution 4.
Solution 4 is very similar to the previous one, but it is index-
ing differently: point M will symmetrically map in relation to
the BC law. So, we get the point K. Now, ]ACK = ]ACM +
]MCB+]BCK = 60◦
(since ]BCK = ]BCM = 10◦
) and ]BKC =
]BMC = 180◦
− ]MBC − ]MCB = 140◦
.
In the circle with center A and radius AB, since AB = AC we
have C ∈ k. Let Y be the arbitrary point of the circle k such that
the points Y and K are made from different sides by BC. Then,
we have 1
2 ]BYC = ]BAC = 40◦
, So, ]BYC + ]BCC = 180◦
,
which means that four-legged CYBK is tetive. Since the points C,
Y, and B belong to the circle k we have that K ∈ k. Now, ]BAK =
2]BCK = 20◦
, so ]KAC = ]BAC − ]BAK = 80◦
− 20◦
= 60◦
.
Since we have already shown that ]ACK = 60◦
it follows that the
AKC triangle is equidistant, hence, AC = KC. Since the points M
and K are symmetric in relation to BC, we have MC = KC, so AC
= MC. We have again proved that the triangle AMC is equivalent
when ]AMC = ]MAC = 70◦
, completes solution 4.
Solution 5
Solution 5 is trigonometric as the sinusoid theorem is used along
with trigonometric identities.
Let P, Q, and R be the norms from point M, respectively to
pages BC, AC and AB in a row.
We mark AM along with x, and we have ]BAM = ϕ. Now, MR
= AM since sin ϕ = x sin ϕ i MQ = AM sin (80◦
– ϕ) = x sin
(80◦
– ϕ). From the triangle BMR we have
BM =
MR
sin20◦
, hence, BM =
xsinϕ
sin20◦
.
From the CMQ triangle, CM = MQ
sin40◦ , hence, CM = xsin(80− ϕ
sin40◦ .](https://image.slidesharecdn.com/article1ijscit-210127093107/85/Different-Ways-of-Solving-a-Geometric-Task-2-320.jpg)
![Different Ways of Solving a Geometric Task 3
Now, applying the sinus theorem to the BCM triangle we have
BM
sin10◦ = CM
sin30◦ i.e., xsinϕ
sin10◦sin20◦ = xsin(80− ϕ
sin30◦ sin40◦ .
Hence, 1
2 sin ϕ sin 40◦
sin(80◦
− ϕ) ) sin 20◦
sin10◦
. The fol-
lowing account is one of the ways to solve the previoustrigono-
metricequation:
1
2
(cos(ϕ−40◦
)−cos(ϕ+40◦
)) = (cos(70◦
−ϕ) cos(90◦
−ϕ)) sin 20◦
1
2
(cos(ϕ−40◦
)−cos(ϕ+40◦
)) = sin(20◦
+ϕ) sin 20◦
−sin ϕ sin 20◦
1
2
(cos(ϕ − 40◦
) −
1
2
cos(ϕ + 40◦
) =
1
2
cos ϕ −
1
2
cos(ϕ + 40◦
)
− sin ϕ sin 20◦
2 sin ϕ sin 20◦
= cos ϕ − cos(ϕ − 40◦
)
2 sin ϕ sin 20◦
= −2 sin(ϕ − 20◦
) sin 20◦
sin ϕ + sin(ϕ − 20◦
) = 0
2 sin(ϕ − 10◦
) cos 10◦
= 0
Since 0 <ϕ <80◦
we have sin(ϕ − 10◦
) = 0 and hence ϕ = 10◦
.
Now, ]MAC = 80◦
− ϕ = 70◦
and ]AMC = 180◦
− ]MAC −
]ACM = 70◦
.
This completes solution 5.
Now, we provide a solution through complex numbers also.
As in analytical geometry wherein we use arranged pairs of real
numbers as coordinates of the point, we canuse the complex num-
ber z = x + i y as coordinate of the point, since the complex
number z can be identified with z arranged steam (x, y). In this
regard, we mention two variants of the solution: in the first we
show that AC = CM, as it is similar to ]AMC, and we discuss
another corner to work out the proof directly.
CONCLUSION
The importance of mathematics for civilization in general is very
important. Mathematics is applied in other sciences, art, and cul-
ture. The progress of mankind is unthinkable without using the
results of natural sciences, the basis of which is mathematics
which provides methods and tools for further research and im-
provement.
The practical importance of mathematics is well known. Fur-
ther, the significance of mathematics is endorsed in our educa-
tional system. Sometimes, we feel that mathematics is difficult,
and even unnecessary in many spheres. That’s why children, of-
ten before going to school, are scared due to such comments in
their surroundings. Based on the insights provided in our paper,
it’s not our goal to ask the culprits to refrain from spreading such
invalid comments but rather ask them to participate in activi-
ties that demystify such negative opinions and continuosly affirm
positive attitudes toward mathematics and learning mathematics
in both letter and spirit.
References
[1] Vojislav Andric, Solving the problem by differentiating the case, Mathemat-
ics, 3/1981, Školskaknjiga, Zagreb, 1981.
[2] Vojislav Andric, Diophantine Equation, Society of Mathematicians of Ser-
bia, Valjevo, 2008.
[3] MaricaPrešic, Method of proving by differentiation of cases, Mathematics,
3/1979, Školskaknjiga, Zagreb, 1979.
[4] ZdravkoKurnik, Special Methods for Solving Mathematical Problems, Ele-
ment, Zagreb, 2010.
[5] Gycev, A.I. Oplov, D.L. Rozental, Extra-curricular Activities in Math-
ematics, Prosvešenie, Moscow, 1984.](https://image.slidesharecdn.com/article1ijscit-210127093107/85/Different-Ways-of-Solving-a-Geometric-Task-3-320.jpg)
Different Ways of Solving a Geometric Task
- 1. International Journal of Smart Computing and Information Technology 2020, Vol. 1, No. 1, pp. 1–3 Copyright c 2020 by BOHR Publishers www.bohrpub.com Different Ways of Solving a Geometric Task Dragan Obradovic1 and Lakshmi Narayan Mishra2∗ 1 Elementary school “Jovan Cvijic”, Kostolac-Pozarevac, Serbia 2 Department of Mathematics, School of Advanced Sciences, Vellore Institute of Technology (VIT) University, Vellore, Tamil Nadu, India *Corresponding author. E-mail: lakshminarayanmishra04@gmail.com Abstract. We present the solution in several ways, i.e., one solution from geometry, four solutions are clearly geometric, i.e., one is trigonometric and three are analytical (two on complex level and one in Deck level). Keywords: Geometry, triangle, circle, equal-angle triangle, trigonometry, task, sinusoid theorem, coordinates, complex num- bers, trigonometric identities Task and solutions Here are some solutions to the following task: In an equal-angle triangle ABC with the base BC the angle at the top A is equal to 80◦ . In the interior of the ABC triangle, the point M is such that the angle ]MBC = 30◦ and angle ]MCB = 10◦ . Find the angle of AMC. Solution 1. Let D be the height of top A on the base of the BC of triangle ABC and let the point N be an intersection of AD and BM. Then there is the angle ]CMN external to the triangle BMC is the same ]CMN = ]MBC + ]MCB = 40◦ . It is also ]NAC = 40◦ because it is half the corner at the top of the equal-triangular triangle ABC. Since N ∈ AD triangle BCN is equal, it is ]NCB = ]NBC = 30◦ . Now, ]NCM = ]NCB − ]MCB = 20 We know that: ]ACN = ]ACB − ]NCB = 20 Now, we have ]MNC = 180◦ − ]NMC − ]NCM = 120◦ i ]ANC = 180◦ − ]NAC − ]NCA = 120 So, the MNC triangle is matched with the ANC triangle based on the following conditions: ]ANC = ]MNC = 120◦ , ]ACN = ]MCN = 20◦ And NC = NC. From this coincidence, we conclude that AC = MC is a triangle ACM equal angle with angle at the top ]ACM = 40◦ . Its base is ]AMC = ]MAC = 70◦ . Thus, we now complete solution 1. Solution 2. Let O be the center of the circle k described around the triangle BMC. As ]MOC = 2]MBC = 60◦ since they are angles ]MOC and ]MBC, respectively central and peripheral over with the MC circle k. Since MO = OC it follows that the triangle MOC is equi- lateral. It is also ]BOM central when it is ]BCM peripheral over with the tendency of the BM circle k, it is ]BOM = 2]BCM = 20◦ . 1
- 2. 2 Dragan Obradovic and Lakshmi Narayan Mishra We now have ]BOC = 80◦ , so since there are basics equal triangles ABC and OBC are equal (that is along BC), as their corners are at the top equal (i.e., ]BAC = ]BOC = 80◦ ) we hence have ABC and OBC matched, so is AC = OC. From the fact that the MOC triangle is equilibrium, we have MC = OC, hence, AC = MC. Thus, we have the AMC triangle with an angle at the top ]MCA = 40◦ . Hence, ]AMC = ]MAC = 70◦ . Thus, solution 2 is now complete. Solution 2 can also be given in the following way, so in the first one the plan comes with the properties of symmetrical fig- ures. Solution 3. Let point O be symmetric with point A in relation to BC law. Then it will be AC = OC and AB = OB, and since AB = AC we have AB = AC = OC = OB. Wehave a circle with the center O and radius OB. Since OB = OC we have C∈k. Let X be an arbitrary point of the circle such that the points X and M are made from different sides by BC. Now, 1 2 ]BXC = ]BOC = 40◦ . Since ]BMC = 180◦ − ]MBC − ]MCB = 140◦ , we have ]BMC + ]BCC = 180◦ , so the BXCM four-blade is tetitive. Now, the points B, X, and C already on the circle k follow that M be- longs to k, so MO = OC. Now, ]MOC = 2]MBC = 60◦ , so the MOC triangle is equilateral, thus it follows that MC = OC. Since we already have AC = OC, it follows that MC = AC, so the MAC triangle is equal, so as in the previous version we have that ]AMC = ]MAC = 70◦ . Solution 4. Solution 4 is very similar to the previous one, but it is index- ing differently: point M will symmetrically map in relation to the BC law. So, we get the point K. Now, ]ACK = ]ACM + ]MCB+]BCK = 60◦ (since ]BCK = ]BCM = 10◦ ) and ]BKC = ]BMC = 180◦ − ]MBC − ]MCB = 140◦ . In the circle with center A and radius AB, since AB = AC we have C ∈ k. Let Y be the arbitrary point of the circle k such that the points Y and K are made from different sides by BC. Then, we have 1 2 ]BYC = ]BAC = 40◦ , So, ]BYC + ]BCC = 180◦ , which means that four-legged CYBK is tetive. Since the points C, Y, and B belong to the circle k we have that K ∈ k. Now, ]BAK = 2]BCK = 20◦ , so ]KAC = ]BAC − ]BAK = 80◦ − 20◦ = 60◦ . Since we have already shown that ]ACK = 60◦ it follows that the AKC triangle is equidistant, hence, AC = KC. Since the points M and K are symmetric in relation to BC, we have MC = KC, so AC = MC. We have again proved that the triangle AMC is equivalent when ]AMC = ]MAC = 70◦ , completes solution 4. Solution 5 Solution 5 is trigonometric as the sinusoid theorem is used along with trigonometric identities. Let P, Q, and R be the norms from point M, respectively to pages BC, AC and AB in a row. We mark AM along with x, and we have ]BAM = ϕ. Now, MR = AM since sin ϕ = x sin ϕ i MQ = AM sin (80◦ – ϕ) = x sin (80◦ – ϕ). From the triangle BMR we have BM = MR sin20◦ , hence, BM = xsinϕ sin20◦ . From the CMQ triangle, CM = MQ sin40◦ , hence, CM = xsin(80− ϕ sin40◦ .
- 3. Different Ways of Solving a Geometric Task 3 Now, applying the sinus theorem to the BCM triangle we have BM sin10◦ = CM sin30◦ i.e., xsinϕ sin10◦sin20◦ = xsin(80− ϕ sin30◦ sin40◦ . Hence, 1 2 sin ϕ sin 40◦ sin(80◦ − ϕ) ) sin 20◦ sin10◦ . The fol- lowing account is one of the ways to solve the previoustrigono- metricequation: 1 2 (cos(ϕ−40◦ )−cos(ϕ+40◦ )) = (cos(70◦ −ϕ) cos(90◦ −ϕ)) sin 20◦ 1 2 (cos(ϕ−40◦ )−cos(ϕ+40◦ )) = sin(20◦ +ϕ) sin 20◦ −sin ϕ sin 20◦ 1 2 (cos(ϕ − 40◦ ) − 1 2 cos(ϕ + 40◦ ) = 1 2 cos ϕ − 1 2 cos(ϕ + 40◦ ) − sin ϕ sin 20◦ 2 sin ϕ sin 20◦ = cos ϕ − cos(ϕ − 40◦ ) 2 sin ϕ sin 20◦ = −2 sin(ϕ − 20◦ ) sin 20◦ sin ϕ + sin(ϕ − 20◦ ) = 0 2 sin(ϕ − 10◦ ) cos 10◦ = 0 Since 0 <ϕ <80◦ we have sin(ϕ − 10◦ ) = 0 and hence ϕ = 10◦ . Now, ]MAC = 80◦ − ϕ = 70◦ and ]AMC = 180◦ − ]MAC − ]ACM = 70◦ . This completes solution 5. Now, we provide a solution through complex numbers also. As in analytical geometry wherein we use arranged pairs of real numbers as coordinates of the point, we canuse the complex num- ber z = x + i y as coordinate of the point, since the complex number z can be identified with z arranged steam (x, y). In this regard, we mention two variants of the solution: in the first we show that AC = CM, as it is similar to ]AMC, and we discuss another corner to work out the proof directly. CONCLUSION The importance of mathematics for civilization in general is very important. Mathematics is applied in other sciences, art, and cul- ture. The progress of mankind is unthinkable without using the results of natural sciences, the basis of which is mathematics which provides methods and tools for further research and im- provement. The practical importance of mathematics is well known. Fur- ther, the significance of mathematics is endorsed in our educa- tional system. Sometimes, we feel that mathematics is difficult, and even unnecessary in many spheres. That’s why children, of- ten before going to school, are scared due to such comments in their surroundings. Based on the insights provided in our paper, it’s not our goal to ask the culprits to refrain from spreading such invalid comments but rather ask them to participate in activi- ties that demystify such negative opinions and continuosly affirm positive attitudes toward mathematics and learning mathematics in both letter and spirit. References [1] Vojislav Andric, Solving the problem by differentiating the case, Mathemat- ics, 3/1981, Školskaknjiga, Zagreb, 1981. [2] Vojislav Andric, Diophantine Equation, Society of Mathematicians of Ser- bia, Valjevo, 2008. [3] MaricaPrešic, Method of proving by differentiation of cases, Mathematics, 3/1979, Školskaknjiga, Zagreb, 1979. [4] ZdravkoKurnik, Special Methods for Solving Mathematical Problems, Ele- ment, Zagreb, 2010. [5] Gycev, A.I. Oplov, D.L. Rozental, Extra-curricular Activities in Math- ematics, Prosvešenie, Moscow, 1984.