A Nice Geometry Problem | You should know this trick !!

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A Nice Geometry Problem | You should know this trick !!
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Комментарии : 41

@planetwazza Год назад

Couldn't you just igmore the trigonometry and say by Pythagorus x squared plus y squared = 20 squared = 400 and sloved this with fewer steps?

@jimlocke9320 Год назад

Staying "inside the box": At 7:20, consider only the constructions that are inside ΔABC. Note that ΔCB₃F, ΔCB₂E, ΔCB₁D and ΔCBA are similar by angle-angle ( the common

@vacuumcarexpo Год назад

The center of the circumscribed circle of a right triangle locates in the middle of the hypotenuse of the right triangle. Therefore, b=AE=EC=40. By the parallelogram law, AB^2+b^2=2(a^2+20^2)…① BC^2+b^2=2(c^2+20^2)…② By Pythagorean Theorem, AB^2+BC^2=80^2…③ (①+②-③)÷2 b^2=a^2+c^2-2400 ⇔a^2+c^2=4000(∵b=40) ⇒a^2+b^2+c^2=5600.

@amanueltesfu-hu2eu Год назад

Used Stewart's Theorem and got the same answer

@prime423 5 месяцев назад

The Median to the hypotenuse of a right triangle is 1/2 the hypotenuse. Use stuarts formula for the median to finish this problem.

@jimlocke9320 Год назад

In this problem, the lengths of AB and BC are not given. The problem statement implies that all valid triangles must produce the same solution. So, I look for a special case which will be straightforward to solve. Let's choose AB = BC, where ΔABC is a special isosceles 45°45°90° right triangle. By symmetry, we find that ΔAEB and ΔCEB are also special isosceles 45°45°90° right triangles, so length EB = b = 40.We note that the squares of length BD = a and BF = c are, by Pythagorean theorem, equal to 40² + 20² = 2000. b = 40 so b² = 1600. a² + b² + c² = 2000 + 1600 + 2000 = 5600. Another special case is the limiting case as AB becomes infinitesimally small. Then a = 20, b = 40 and c = 60. So a² + b² + c² = 20² + 40² + 60² = 400 + 1600 + 3600 = 5600. The problem statement implies that the solution for any valid pair of values for AB and BC applies to all valid pairs, so we can argue that we solved the problem. To thwart us, the problem statement could be rewritten as follows "Prove that a² + b² + c² is the same for all valid ΔABC and provide the value of a² + b² + c²."

@tunneloflight 10 месяцев назад

I had the same thought. Unsupported but inferred, the answer is either that all such triangles result in the same sum and that is the answer, or they don’t and the answer to the question is indeterminant (or an equation). Then proceeding as you described I reached the same conclusion. The only difference is that I started with the reduced case of a line (one side vanishingly close to zero height). Then proceed to the 45 degree angle case you started with. The result is the same. Confidence is then high. It is easy to extend that to the special case of the 30-60-90 right triangle. That too results in the same answer. Confidence is now certain. The only advantage in doing this is in reaching the answer more quickly to allow moving on to the next question. Otherwise, the rigorous solution proceeds through several paths just has he describes to then analytically prove that all cases result in the same answer. But that wasn’t the problem statement. Proof of all cases wasn’t required to be demonstrated, only the answer to the sum of the three lengths. So the reduced case is justified.

@zsboya Год назад

More simply: x = 20 unit , - Thales-circle: AC = 2r = 4x -> r = AE = CE = BE = AC/2 -> b = 2x -> b^2 = 4*x^2 , - DBF triangle, by Stewart's Theorem: x*(a^2 + c^2) = 2x*(b^2 + x^2) -> a^2 + c^2 = 2*(4*x^2 + x^2) = 10*x^2 -> sum(s^2) = a^2 + c^2 + b^2 = (10+4)*x^2 = 14*20^2 = 5600 unit^2 . \ -Note: s(s^2) = (2*(2^2+1^2) + 2^2) * x^2 = (3*2^2 + 2*1^2) * x^2 = 14*x^2 \ Or: - DBF triangle, by geometric median: m(DF) = b -> (2*m(DF))^2 = (2b)^2 = 2*(BD^2 + BF^2) - DF^2 = 2*(a^2 + c^2) - (2x)^2 -> a^2 + c^2 = ((2b)^2 + (2x)^2)/2 = 2*(b^2 + x^2) -> /+ b^2 -> s(s^2) = a^2 + b^2 + c^2 = 3*b^2 + 2*x^2 = (12+2)*x^2 = 14*20^2 = 5600 unit^2 . Or: - DBF Δ -> DBFB' parallelogram, by Paral. law ( 2*(s1^2 + s2^2) = d1^2 + d2^2 ): sides: a, c & diagonals: 2b, 2x -> a^2 + c^2 = ((2b)^2 + (2x)^2)/2 = 2*(b^2 + x^2) -> ...

@HERKELMERKEL 8 месяцев назад

you struggle too much with the limitations of youtube comments section 😅 they definitely must add text formatting, or at least latex commands so it can be readable easily.

@arantheo8607 Год назад

Thank you for presenting a different approach to the problem. The sum of the squares of any two sides of any triangle equals twice the square on half of the third side, plus twice the square on the median bisecting the third side a is mediam in triangle ABE b is mediamin triangle ABC c is mediam in triangle BCE Also, In a right triangle, the length of the median drawn from the vertex of the right angle equals half the length of the triangle's hypotenuse, therefore b = 40 applying Apollonius Theorem and rewrinting in a more convenient form a² = [2(AB)² +2(BE)² - 40²]/4 b²= [2(AB)² +2(BC)² - 80²]/4 c² = [2(BE)² +2(BC)² - 40²]/4 a²+b²+c² = [2(AB)² +2(BE)² - 40²]/4 +[2(AB)² +2(BC)² - 80²]/4+ [2(BE)² +2(BC)² - 40²]/4 a²+b²+c² = [4(AB)² +4(BE)² +4(BC)² - 40²-40²-80²]/4 a²+b²+c² = [4[ (AB)² +(BC)² ] +4(BE)² - 2.40²- 4.40²]/4 a²+b²+c² = [4 (80)² +4(40)² - 6.40²]/4 a²+b²+c² = [16. 40² +4(40)² - 6.40²]/4 a²+b²+c² = [16 +4 - 6 ).40²]/4 a²+b²+c² = 14.( 1600)/4 a²+b²+c² = 5600

@edsznyter1437 5 месяцев назад

Draw horizontal and vertical lines through D, E, and F. This forms a regular grid, which breaks up each leg of the big triangle into four equal segments. The little triangle by CF gives us x^2 + y^2 = 20^2. The lines a, b, and c are diagonals of rectangles in the grid, and a^2+b^2+c^2 = [(3x)^2+y^2] + [(2x)^2+(2y)^2] + [x^2+(3y)^2] = 14(x^2+y^2) = 14*20^2 = 5600. This is a do-in-your-head problem. (Restatement of @jimlocke9320's)

@hongningsuen1348 7 месяцев назад

I would like to use the simple intercept theorem to explain your solution method: 1. Draw 2 sets of parallel lines, one vertical set parallel to AB and one horizontal set parallel to BC, connecting D, E, F. 2. By intercept theorem, AB is divided into 4 equal vertical segments. Let it be v. Similarly by intercept theorem, BC is divided into 4 equal horizontal segments. Let it be h. 3. By Pythagoras theorem: a^2 = h^2 + (3v)^2 b^2 = (2h)^2 + (2v)^2 c^2 = (3h)^2 + v^2 Hence a^2 + b^2 + c^2 = 14 x h^2 + 14 x v^2 = 14 (h^2 + v^2) = 14 x 20^2 = 5600.

@jehmacapagal3804 11 месяцев назад

Since F is the midpoint of AC, then BF=b = 40. Since DF is half of BF, BDF is a right triangle. a= 20√3. By Pythagorean Theorem, c²= 40²+1200=1600+1200=2800 Hence, a²+b²+c²=1600+1200+2800=5600

@davidbennett1316 8 месяцев назад

cosine rule also seems to deliver solution quite neatly applying successively to ADB, AEB, AFB and for ACB it is simply cosA= AB/80 - equate all these and simplify you get 2a^2 = 2.20^2 + AB^2, 2b^2 = 2.40^2 and 2.c^2 = 2.60^2 - AB^2 which quickly => 5,600

@petersteiner9820 Год назад

Nice. Another approach to find that b=40, is that if AC is the Diameter than B lies on the Thales circle and E is the center of the circle so AE=EC=EB=r radius of the circle which is 40.

@alexandertokarev2097 9 месяцев назад

Для данной задачи удобно воспользоваться геометрией масс. Пусть гипотенуза разделена на n равных отрезка длины l. Поместим в n-1 точек, разделяющих отрезки грузы массой 1. Момент инерции относительно вершины прямого угла C равен искомому значению JC=CD^2+CE^2+CF^2+... Центр масс системы - в середине гипотенузы AB. В соответствии с теоремой Штейнера момент инерции равен JC для любой точки окружности (Штейнера) с центром в середине гипотенузы и проходящей через C. Эта же окружность проходит через A и B. Найдем искомый момент инерции относительно A (C и B), воспользовавшись формулой суммы квадратов натуральных чисел: JC=JA=JB=l^2+(2•l)^2+(3•l)^2+... +(n-1)^2•l^2 = (n-1)•n•(2n-1)•l^2/6.

@MarcosHeitorCardoso Год назад

b= 40, median of the triangle. Applying two stewart's theorem to 2 triangles for a e c e adding both equations the a2+b2 = 4000. Plus b2=1600, we arrived at 5600. However, I have liked your solution more. Best!

@nishantshelar8887 11 месяцев назад

Use apollonius's theorem, to get 4 equations and then solve for the sum.

@arafatalleion6765 8 месяцев назад

We can also use Stewart's Theorem. If we use it, we will get the same answer

@iossifid 11 месяцев назад

Another solution is the following: b is median of right triangle ABC, so b=AC/2=40, so b^2=1600 According to the 1st theorem of medians, from the triangle BDF we have: a^2+c^2=2b^2+(DF^2)/2=2.40^2+(40^2)/2=3200+800= 4000 Hence a^2+b^2+c^2=4000+1600=5600

@kaliprasadguru1921 Год назад

Drawing DB1 perpendicular to BC is not correct . A1D and DB1 may not lie in one line . Hence you have to extend A1D to B1 so that AA1B1B will be a rectangle and then AA1=BB1=x . Regards

@Bossudeboss898 11 месяцев назад

It is as correct as it has to be. DB1 || AB because they are perpendicular to a secant (BC). Also DA1 || AB because they are perpendicular to a secant (AM). But from the axiom of a parallel > B1D1 || AB, hence the conclusion.

@KSM94K Год назад

It can be more simply solved(completely in my mind but I think I didn't do any mistake) a²=(AB-AB/4)²+AC²/16 b²=(AB-AB/2)²+AC²/4 c²=(AB-3AB/4)²+9AC²/16 [After drawing perpendiculars and from triangle similarities we found the components of the hypotenuse and showed it in terms of AB and AC] Now just add them and you know AB²+BC², and some terms would become common and you would get the answer

@newzero1000 Год назад

I did same way. you had a typo here though: AC should be BC here. (1+4+9)/16*(4*20)^2 = 5600. This is much easy way. a²=(AB-AB/4)²+BC²/16 b²=(AB-AB/2)²+BC²/4 c²=(AB-3AB/4)²+BC²/16 add them together: (1+4+9)/16*(AB^2 + BC ^2) = (1+4+9)/16*(AC)^2 = (1+4+9)/16*(4*20)^2 = 5600 further more. for the general case: if AC is divided n pieces with each length = m. the solution will be: (n-1)n(2n-1)/6*m^2. For this special case, n = 4 and m = 20, the answer is 5600.

@KSM94K Год назад

@@newzero1000 yes you're right, that's BC And I probably didn't check the last expression, it's just dividing a L length into n parts and each of them are L/n, a1²=(AB-AB/n)²+BC²/n²=AB²(1-1/n)²+BC/n² a2²=AB²(1-2/n)²+4BC²/n² a3²=AB²(1-3/n)²+9BC²/n² .......a(n-1)²=AB²[1-(n-1)/n]²+(n-1)²BC²/n² Just add them and if think there's a series pattern in this or maybe my calculations are incorrect

@zsboya Год назад

More simply: x = 20 unit , Let the right triangle ABC be an isosceles triangle: y = AB = BC (45°-45°). Then the segments a, b, c divide the angle Due to the Thales circle, r = AE = EC = BE -> b = 2x, and BE is perpendicular to AC -> BEDΔ and BEFΔ are congruent right triangles. -> By Pythagoras' theorem: a^2 = c^2 = b^2 + x^2 = 5*x^2 -> sum(s^2) = 2*a^2 + b^2 = (10+4)*x^2 = 14*20^2 = 5600 unit^2 .

@davidellis1929 Год назад

Here's a slight simplification of your final step. From the small right triangles after your constructions, x^2+y^2=20^2=400. You don't need any trigonometric functions to solve the problem.

@sunson-r1d 11 месяцев назад

nice solution

@lasalleman6792 Год назад

Pretty elegant. And pretty hard. I couldn't figure it out. Had to check the solution. Good exercise thouigh.

@PrithwirajSen-nj6qq 3 месяца назад

Bravo!

@RAG981 Год назад

I would consider that it equals 7/8of AB^2 to be the interesting thing.

@AAZ3000 Год назад

Sir, how much time did you spent before solving the problem this way?

@محمدنبي-د5ف Год назад

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

@SundalKlakson Год назад

Velmi jednoduchá úloha. Ze zadání je jasné, že řešení nezávisí na velikosti úhlu BCA. Pokud se jeho velikost blíží nule, platí: a = 20, b = 40 a c = 60. ... (A very simple task. It is clear from the assignment that the solution does not depend on the size of the angle BCA. If its size is close to zero, then: a = 20, b = 40 and c = 60.)

@yuusufliibaan1380 Год назад

❤❤ thanks 🙏🙏 my dear teacher ❤️💯👍

@howardaltman7212 Год назад

Very creative approach professor.

@kkmaheshwari568 Год назад

EXCELLENT PRESENTATION

@kkmaheshwari568 11 месяцев назад

@@English_shahriar1 By using Appolonius theorem...it can be solved

@sumithpeiris8440 11 месяцев назад

2 minutes solution E is the centre of Right Triangle ABC so b = 40 Now apply Appolonius Theorem to Triangle DBF a^2 + c^2 = 2b^2 + 2 X 20^2 = 4,000 Hence a^2 + b^2 + c^2 = 4,000 + 40^2 = 5,600