A Nice Math Olympiad Geometry Problem | 2 Methods to Solve

[@georgexomeritakis2793]

There is also a pure geometrical solution to this puzzle. Extent AC downwards and take a point E so that you create an isosceles triangle BEA with angles 50,80,50. This way the triangle BEC is also isosceles with angles 20,80,80. Now extent BC to the right and take a point F so that triangle BEF is isosceles. This will have angles 20,140,20. Now you going to have BE=AE=BC=EF. Now notice that angle AEF will be 60 = 140-80, so the triangle AEF will be equilateral. Now angle AFC will be 60-20 = 40. So since AD=AF, then the unknown angle will be also 40. Sorry I cannot provide a graph in the comments....

[@РешебникЕГЭОГЭМатематика]

Здравствуйте, Георгий! Все отлично, никаких чертежей не нужно! Мое решение несколько иное и основано на вписанных углах и четвертом признаке равенстве треугольников. Так что ваше решение более понятно и проще. Спасибо. Γεια σου Γεώργιο! Όλα είναι υπέροχα, δεν χρειάζονται σχέδια! Η λύση μου είναι ελαφρώς διαφορετική και βασίζεται σε εγγεγραμμένες γωνίες και στο τέταρτο κριτήριο για την ισότητα των τριγώνων. Άρα η λύση σας είναι πιο ξεκάθαρη και πιο απλή. Ευχαριστώ. Hello, Georgy! Everything is great, no drawings needed! My solution is slightly different and is based on inscribed angles and the fourth criterion for the equality of triangles. So your solution is more clear and simpler. Thank you.

[@xunningyue9901]

Elegant solution. Mine is different from you at pt F, where my F p.t. is the symmetrical pt of A about line BC. Since AB=FB, angle ABF=2angle ABC = 60, we get equilateral ABF hence AF = AB. Since AB=FB=BE=AE, we have triangle ABE congruent to AFD, thus angle theta = angle ADF/2 = angle AEB/2 = 40.

[@SuperPassek]

Awesome!

[@DB-lg5sq]

شكرا على مجهوداتكم

[@jinyu500]

amazing

[@DB-lg5sq]

شكرا على المجهودات يمكن تبسيط واختصار الطريقة بأخذ BC=1 ........ نجد x=40

[@grzegorzmordas9134]

Angles in this triangle: 30, 100, 50 deg. Name side opposite 30 deg.: c, side opposite 100 deg.: b, side opposite 50 deg.: a. Then from the sines formula: a/sin50 = b/sin100 and thus: b/a = sin100/sin50. We can also write: a*sin(Theta) = b*sin30, which means that: sin(Theta) = b*sin(30)/a, therefore: sin(Theta) = sin(100)*sin(30)/sin(50) = 0.6427876.... arc sin 0.6427876 = 40 deg.

[@Latronibus]

Strictly speaking you should write sin(100) sin(30)/sin(50)=cos(50)=sin(40) instead of assuming your calculator result is exact.

[@alishernizamov9441]

АС/sin30 =BC/sin50 and AC/sinx = AD/sin100 ,(AD=BC), x=40

[@SushilKumar-ds8et]

😊😊😊😊😊😊😊😊😊😊😊

[@MarieAnne.]

I used the same method you did (in Method 1), but since we don't care about ∠BAD and ∠DAC, I don't see the usefulness of splitting ∠BAC into these two angles. Instead, I just let θ = ∠ADC (since that's the angle we need to find). So using law of sines in △ABC we get a/b = sin A / sin B = sin 50 / sin 30 and using law of sines in △DAC we get a/b = sin C / sin D = sin 100 / sin θ This gives us: sin 50 / sin 30 = sin 100 / sin θ sin θ = sin 30 sin 100 / sin 50 = 1/2 * 2 sin 50 cos 50 / sin 50 = cos 50 = sin(90-50) = sin 40 ∠ADC = θ = 40

[@alegoncalves472]

Amazing!!!

[@Ivan-Matematyk]

A short, purely geometrical solution. Let O be the center of the circumscribed circle of the triangle ABC and the point E on the BC side such that CE=CA. Since the central angle BOC is twice the inscribed angle BAC, it is equal to 100 degrees. Similarly the central angle COA is twice the inscribed angle ABC and it is equal to 60 degrees. Given also the equality OA=OC, we get that the triangle OAC is equilateral. Therefore, BO=OC=AC=CE. Thus, the triangles BOC and ACE are equal and AE=BC. If, in this case, D and E are different points, then AD is not equal to AE, and therefore is not equal to BC. It follows that the angle ADC is equal to the angle of AEC and is equal to 40 degrees.

[@jasperbear2248]

Very nice. Impossible to think of such a clever solution!

[@jimlocke9320]

An observation: At 15:30, we have sin(Θ) = (tan(80°))/((√3)tan(80°) - 1). Unless we need an exact solution, the problem is solved because can calculate Θ using a scientific calculator and find that Θ = 40° to within the precision of our calculator. If we needed higher precision, we could go to the infinite series formulas for the trigonometric functions and compute to any required level of precision, although the higher the precision the longer it would take to complete the calculations. (The calculation problem is complicated by stating the angles in degrees when the infinite series use radians. However, we can compute π to any required level of precision for the conversion to radians and that precision figures in to the overall precision.) Math Booster's next calculations effectively prove that Θ is exactly 40° and other commentors have provided additional ways to produce an exact value of 40°. Of course, obtaining this very nice result required that the problem be constructed such that Θ is exactly 40°. The calculation method will always produce 40° to within the precision of our calculation. It just won't prove that Θ is exactly 40°! For practical applications, dimensions always have tolerances and engineers just need to compute the result to within the desired tolerance.

[@User-jr7vf]

In a university admission exam they would probably allow for the answer to be given as (tan(80°))/((√3)tan(80°) - 1), but the objective of this channel is to go deep into the math that is presented here, hence why she takes the additional step of evaluating that thing to get 40°.

[@pedromaducdoc415]

I just used 2 eqs consider tri ABC: B=30;C=100;A=50:Let AD=BC=x. Then by sin law BC/sinA =AC/sinB. Solving for AC = xsin30/sin50. Then consider triADC:::AD =x:D=theta:C=100: by sin law AD/sin100 =AC/sintheta: substitute AC and solve for theta. Theta =[sin30sin100/sin50]. Theta = 40

[@اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم جميعا . تحياتنا لكم من غزة فلسطين

[@SushilKumar-ds8et]

😊😊😊😊😊😊😊😊😊😊😊

[@mordechaimalka]

I don’t understand the relation with gaza and Palestine ? Are you crazy or what ?

[@vladimirstoimenov6945]

The most elegant solution: Make the triangleABC right-angled (as at 10:13 minute of this video), namely ABE. The new angle in A is now equal 60°, so angle

[@User-jr7vf]

Wow, very interesting solution. Also the quickest one. You are brilliant for bringing it up . However I would say method 2 of "Math Booster" is the most elegant solution, because of all the math steps that she takes to arrive at the answer. That is to say, it requires a lot of brain.

[@jayeshkumar3861]

Awesome.....

[@vladimirstoimenov6945]

Another solution: If you make the triangleABC right-angled (as at 10:13 minute of this video), namely ABE, you can achieve the solution sinθ = 1/(tan 60° - tan10°). θ = 40°.

[@howardaltman7212]

Thanks for boosting my math over the past few months. I came up with another method where angle ADC = theta and AC=x. Since triangles ABC and ADC are both obtuse, I used the Law of Sines on each without worry. Since each proportion contained (a/x), I could eliminate it through substitution which only left theta. Thanks again for your daily math videos.

[@TheDHemant]

Hats off to you Sir

[@TheDHemant]

You are great Sir

[@hcgreier6037]

I acually doubt that this is an Olympiad question. Let angle CDA = x. The law of sines give 1. AC/sin(x) = AD/sin(100) ⇒ AC·sin(100) = AD·sin(x) 2. АС/sin(30) = BC/sin(50) ⇒ AC·sin(50) = BC·sin(30) Deviding 1. by 2. with AD = BC one gets sin(100)/sin(50) = sin(x)·sin(30) sin(x) = sin(100)·sin(30)/sin(50) Now we can manipulate the sin/cos of angles sin(30) = 1/2 sin(100) = sin(2·50) = 2·sin(50)·cos(50) sin(x) = cos(90-x) one gets cos(90-x) = 2·sin(50)·cos(50)·(1/2)/sin(50) = cos(50) 90-x = 50 ⇒ x = 40 Pretty straight forward.

[@misterenter-iz7rz]

s/sin 100=AC/sin theta, AC/sin 30=s/sin 50, so sin 100 /sin theta=sin 50/sin 30, so sin theta =sin 100 sin 30/sin 50, then theta=40.

[@yuusufliibaan1380]

Thanks dear teacher ❤❤❤

[@giuseppemalaguti435]

Usando il teorema dei seni e del coseno ottengo una cubica in sinT...4(ctg100+ctg30)(sinT)°3-4(sinT)^2-(ctg100+ctg30)sinT+1=0....una prima soluzione è sinT=0.64...non so,dovrei ricontrollare i calcoli...un altro risultato è per sinT=0,696877..(T=44,177)...sinT=0,896...T=63,6488(forse questo sembra il migliore)

[@laxmansingh-en8yr]

by applying appollonius therem it will be QED

[@Cdictator]

Thanks bro. The second method is dopey 😢

[@MT-gt6nc]

Use Alkashi Theorem

[@vijayannair2316]

Nice

[@vijayannair2316]

Thanks

[@РешебникЕГЭОГЭМатематика]

Здравствуйте! Спасибо за канал! Очень интересный ! Но хотелось бы узнать: как решить эту задачу без тригонометрии? Hello! Thanks for the channel! Very interesting ! But I would like to know: how to solve this problem without trigonometry?

[@Ivan-Matematyk]

Наприклад, так: Let O be the center of the circumscribed circle of the triangle ABC and the point E on the BC side such that CE=CA. Since the central angle BOC is twice the inscribed angle BAC, it is equal to 100 degrees. Similarly the central angle COA is twice the inscribed angle ABC and it is equal to 60 degrees. Given also the equality OA=OC, we get that the triangle OAC is equilateral. Therefore, BO=OC=AC=CE. Thus, the triangles BOC and ACE are equal and AE=BC. If, in this case, D and E are different points, then AD is not equal to AE, and therefore is not equal to BC. It follows that the angle ADC is equal to the angle of AEC and is equal to 40 degrees. Нехай O - центр описаного кола трикутника ABC і точка E на стороні BC така , щоt CE=CA. Оскільки центральний кут BOC вдвічі більший вписаного кута BAC, то він дорівнює 100 градусів. Аналогічно центральний кут COA вдвічі більший вписаного кута ABC і він дорівнює 60 градусів. Враховуючи також рівність OA=OC, ми отримуємо, що трикутник OAC є рівностороннім. Тому BO=OC=AC=CE. Таким чином, трикутники BOC та ACE рівні і AE=BC. Якщо при цьому D and E - різні точки, то AD не дорівнює AE і тому не дорівнює BC. Звідси випливає, що кут ADC дорівнює куту AEC і дорівнює 40 градусів.

[@vladimirstoimenov6945]

Make the triangle ABC right-angled (as at 10:13 minute of this video), namely ABE. The new angle in A is now equal 60°, so angle

[@frosinabrahja1792]

B+D+C=105

[@РешебникЕГЭОГЭМатематика]

У меня есть решение Но, возможно, ваше проще и интереснее. Спасибо! I have a solution, but perhaps yours is simpler and more interesting. Thank you!

[@said14121]

رائع ❤❤❤

[@padraiggluck2980]

👍

[@valisahranavard]

با درود وسپاس بسیار

[@SamuelRajThomas]

60 degres

[@LuisSayago-ec3hq]

There's a THIRD EASYS METHOD for the triangle ABC with angles ABC = 30 and angle ACB=100.JUST DRAW THIS TRIANGLE AT SCALE AND BY USING A PROTACTOR THE ANGLE ADC IS 40 DEGREES.....GOT IT???

[@LuisSayago-ec3hq]

Sorry dear but both of your methods are too long and complicated..I did it using only once the sine theorem and I've got the answer easily..

[@bbpatel5152]

60

[@ChangBenjamin]

It's not geometry

[@Domestic_egg]

뭘 이렇게 어렵게 풀어. 한국 고등학생이면 메모지 한 장이면 충분하다.

[@Stevenzhang-ze8ki]

绕来绕去！看的头昏！！