@pickover - squares with side length 1 = 4²
- squares with side length 2 = 3²
- squares with side length 3 = 2²
- squares with side length 4 = 1²
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Total squares. = 1² + 2² + 3² + 4² = 1 + 4 + 9 + 16 = 30
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Margarit Milcea
8.6K posts
Doar era "mediocru"... Cu ghilimele de rigoare, firește! 😄😄😄
Physics In History@PhysInHistory
Fun Fact: Louis Pasteur, a remarkable French scientist who made significant discoveries and inventions in the fields of chemistry, microbiology, and medicine was not a good student in his early years. He failed his first attempt at the entrance exam for the École Normale Supérieure, a prestigious institution for higher education in France. He also struggled with mathematics and physics, and only obtained mediocre grades in his general science degree. 📷unknown
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@sonukg4india h - the height of triangle ABC
sin2a = 2•sin a•cos a
h/8 = 2• (h/4)•√(16-h²)/4
h/8 = (h/2) • √(16-h²) /4
√(16-h²)=1, 16-h²=1, h²= 15, h= √15
16-15 = 1, 64-15 = 49., base of the triangle, BC=7-1=6. Area of the triangle= 6•√15/2 = 3√15
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Albert Einstein visited the Hopi people near the Grand Canyon in 1931, where he was honored with a feathered headdress and a peace pipe at Hopi House. The gesture recognized his pacifist ideals and is preserved in a well-known photograph.
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@CGhineaOficial Da, și Bolojan mai are și viziune: dorește crearea unui pol de centru-dreapta puternic!
m.ziare.com/amp/ilie-boloj…
Problema: cât de stabil ar fi un asemenea pol? Antecedentele, CDR, Alianța DA, ar demonstra că este posibil ce dorește el. CDR mi se pare că a fost cel mai stabil!
Română
"De fapt, acțiunea anti-Bolojan a pus împreună un partid care ar vrea alegeri anticipate, căci e primul în sondaje, și un alt partid, care a crezut că, dând jos guvernul, îl destituie pe premier și din fruntea partidului pe care-l conduce.
Calculul a fost greșit. Ilie Bolojan nu mai e prim-ministru, dar poziția lui în PNL a fost consolidată, după ce 50 din cei 54 de membri ai biroului național au votat pentru trecerea partidului în opoziție.
Mai mult decât atât, Bolojan a căzut de acord cu Fritz ca formațiunile lor să își coordoneze acțiunea parlamentară și politică. E o mutare importantă. Blocul PNL-USR e acum fracțiunea cea mai puternică din camere, având 134 de mandate, cu cinci mai multe decât PSD și cu 43 mai numeroase decât ale AUR."
Cristian Preda, pentru Comunitatea Liberală
comunitatealiberala.ro/unde-dai-si-un…
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@MatildaSpicer We apply the Law of Cosines:
AD² = AB² + BD² - 2AB • BD cos 15 = 8 + 12 - 2•2√2•2√3• cos 15 = 20 - 2•2√2•2√3•(√6/4+√2/4) = 20 - 2•√6(√6+√2) = 20 - 12 - 2√12 = 8 - 4√3 = 4(2 - √3),
AD = 2√(2-√3) = 2(√(3/2) - √(1/2)) = 2(√6/2-√2/2) = √6-√2
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@MatildaSpicer It can also be calculated by directly applying the Law of Cosines to triangle ABD (ABD = 15°). First, we calculate cos 15°:
cos(a-b) = cos(45-30) = cos45 cos30 + sin 45 sin 30 = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4
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Triangle ABC is a right isosceles triangle.
Triangle BCD is a 30-60-90 triangle.
Given that BC = 4, find the length of AD.
#mathpuzzle #Geometry #GeometricProblem
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Ce-i al lui, e al lui! 😀Sunt destui care nu vor să admită că omul e capabil, sau nu le convine ... Cel puțin el a excelat în ceva! În ce a excelat Călin Georgescu? Cel puțin, în matematică, Nicușor Dan a obținut o recunoaștere internațională. Ca președinte? Vom vedea!
Fermat's Library@fermatslibrary
Fun fact: Romania’s president, Nicușor Dan, was one of the 11 students to solve the legendary Problem 6 at the 1988 IMO. He scored a perfect 42/42 that year, alongside future Fields Medalist Ngô Bảo Châu. Ravi Vakil also solved it. Terence Tao, aged just 13, got 1/7 on that problem, aced almost everything else, and still won gold.
Română
@sonukg4india The perpendicular from E to BC passes through the center of the small semicircle, O', of radius r, and intersects the large semicircle at F. EF² = 9/4 – 1/4 = 8/4 = 2, EF = √2. The power of point O' in the large circle: (√2 - r) • (√2 + r) = r², r = 1, x = 2.
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@pickover And speaking of point 3, people can always learn! And, generally speaking, people learn from other people...
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@pickover I’m a little skeptical about point 1, but, anyway... Maybe there are people like that, but how many of them can look at the diagram and, without doing any calculations, accurately estimate that the shaded area is larger than 20 square units—and without fear of being wrong?
😄😄
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Mathematics. There are three kinds of people on Earth:
❄️1) Those who can simply look at this diagram and know if the area of the blue region is greater than 20 units squared.
❄️2) Those who can actually compute the area.
❄️3) Those who can do neither 1 or 2).
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@MatildaSpicer If point A lies at the zenith relative to D on the circumcircle of triangle ABC, let A' be this point; then A'DC = 90°, A'DA = A'DC – ADC = 90° – 45° = 45° = A'BA. And A'BD = ABD + A'BA = 30 + 45 = 75. AD-the perpendicular bisector of |BC|, passing through D and center of circle.
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@MatildaSpicer The angles ABD = 30°, ADB = 135°. In the circle circumscribed around triangle ABC, the arc BAC encompasses the angle BAC=15° + x. When point A lies on the perpendicular bisector of |BC|, triangle ABC is isosceles, and ∠ABC = ∠ACB = 75°. -> 15 + x + 135 = 180, x = 30.
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@MatildaSpicer It is also interesting to note that we can determine all the angles of the trapezoid: ADC = 130, ABC = 80, and, of course, BAD = 100, BCD = 50. 130 + 50 + 80 + 100 = 360.
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@MatildaSpicer AB + AD = 3 + 4 = 7 = BC. Let E be a point on BC such that EC = 3. AECD is a parallelogram; angle EAD = 50° = ECD. Triangle ABE is isosceles; AB = BE = 4; angles BAE and BEA are equal. AE||CD, ECD = BEA = BAE = 50. x = BAD = BAE + EAD = 50 + 50 = 100. x = 100.
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