Use Similar Triangles ABC and AOE , x=r/4 and r=12/7,then x=3/7

@franciscogeorge5879

5 ай бұрын

YES. ITS NOT NECESSARY WHAT SHE DID. SIMILARITY IS SO MUCH EASIER AND QUICKER

@ercsrc

5 ай бұрын

I came to write the same thing and timestamp this onwards 3:57 3-r, r, r+x is 3,4,5 triangle

@markmajkowski9545

4 ай бұрын

By 345 Pythagoras gets the Day Off Unknown U = R/4 you subtract the 4 side from the 5 side to get unknown. 5 = (5/3 + 5/4) R or 1 = (1/3 + 1/4) 4U 1= (4/3 + 3/3) U 1= 7/3. U U =3/7

@RakeshKumarRobot70

4 ай бұрын

Mene bhi yahi Kiya hai😊

@rbbza2749

3 ай бұрын

1min without a Pen. Lol

Wow! Alternatively, OBD is similar to ABC which is a 3-4-5 triangle OD = r, BD = 4-r (4 - r) /r = 4/3 12 - 3r = 4r 7r = 12 r = 12/7 Also OB = (5/3)r = 20/7 x = 5 - OB - r x = 35/7 - 20/7 - 12/7 x = 3/7

@FlavianTicaDeme

5 ай бұрын

This is the shortest way to the solution!

@albertybermudez7454

5 ай бұрын

That is correct

@cosimo7770

5 ай бұрын

Of course, @MrPwmiles has the most elegant solution, obvious within seconds. Once again, a YT 'teacher', Math Booster, lacks mathematical insight and any sense of symmetry.

@MichaelRothwell1

5 ай бұрын

Yes, that's how I did it!

@alitn588

4 ай бұрын

3,4,5 rule its very handy! 👍

At 4:15 you start area calculations. THIS IS EXCESSIVELY COMPLICATED! Solve for r knowing triangle BDO is similar to BCA. r/(4-r) = 3/4. From this 4r = 12-3r & r = 12/7 = OP. OA = (5/4)r = 15/7 X = PA = (15/7) - (12/7) = 3/7. No quadratic equation needed!

Nice question, this can also be solved using similar triangles. Would be faster and easier - dont need to deal with quadratics...

The problem can be solved much simpler. Two equations: 1) (r+x)/r = 5/4 2) (r+x)/(3-r) = 5/3 1) solves to r = 4x Filling in 4x for r in 2) solves to x = 3/7

We don't have to use similarity when finding r. Since ABC=OBC+OAC, 6=2r+3r/2. This is much easier.

@bernardbaz2003

3 ай бұрын

Clever !

draw a horizontal line and a perpendicular line from the center of the circle r / (r + x) = 4/5 5r = 4r + 4x x = r/4 r / (3 - r) = 4/3 3r = 12 - 4r r = 12/7 x = r/4 = 3/7

What a convoluted way to get to that answer!

Невероятная сложность решения в ролике меня просто поразила. :)))))))))) Центр окружности - это основание биссектрисы прямого угла (потому что равноудален от сторон угла :)) ), то есть делит гипотенузу на отрезки 5*3/(3+4)=15/7 и 20/7; Центр, точки касания и вершина прямого угла - это вершины квадрата. За пределами квадрата остаются два треугольника, подобных исходному, с гипотенузами 15/7 и 20/7 (в сумме 5, разумеется), откуда радиус (меньший катет одного треугольника и больший у другого) 12/7 (тут обычно трудности с восприятием этой совершенно элементарной мысли, поэтому поясню - у треугольника с гипотенузой 15/7 и подобного "египетскому" (3,4,5) на самом деле известны и катеты 9/7 и 12/7, а не только гипотенуза :) больший катет как раз и будет радиус этой окружности) Осталось 15/7 - 12/7 = 3/7. Очень сложная задача, ну совсем олимпиадная. :)))

@nickvin3212

4 ай бұрын

Устная задача,из подобия тр след (3-R)/3=R/4 , след R=12/7, тк ОА=15/7(гипотенуза а катеты 12/7 и 9/7) то x=3/7.

Alternate method with simple calculations in coordinate system: center of circle is the intersection of (A,B) and bisectiing line of angle BCA. Choose coordinte center in C. So we have x/4 + y/3 = 1 and y = -x. => O(-12/7;12/7) => r=12/7. Then x = dist(O,A) - r = 3/7. No rocket science, few geometry rules, simple doing even for pupils.

CONGRATULATON!YOU FIND THE MOST CUMBERSOME WAY TO SOLVE THIS PROBLEM!

You did not mention in the question that A B passes by the center of the circle. Put the origine at C, the center of the circle is (-r , r). The line AB has the equation 4y - 3x = 12. So 7r = 12. Now x = SQRT (sqr(r) + sqr(3-r)) - r. Replace r by 12/7 and you got the answer.

There is no O in the center of semi circle at the start of this task. Where did you get that? 'O' can be elsewhere..

@umarhassan1807

3 ай бұрын

And OP = r on what basis?

@SGR-fr7hp

2 ай бұрын

He's wrong.

Let's consider the triangles OAC and OBC then solve the equation Area (ABC)=Area(Oac)+ area(obc) to find the radius Then Pythagoras theorem to calculate x

The main triangle is a classic 3-4-5 right triangle. Its easier to use side ratio. Exc triangle OEA OE:4::EA:3, then r/4=(3-r)/3 there you get r=12/7 And then OA:5::OE:4, then (r+x)/5=r/4, then 4r+4x=5r, then 4x=r, so x=r/4=3/7... done in less than 1 minute without hassle

can be easier using 3-4-5 triangle (and previously determining r = 12/7) AB = x + r + OB (with OB = sqrt(r²+(4-r)²) from Pythagorus) => 5 = x+ r + sqrt(r²+(4-r)²) => x = 5 - r - sqrt(r²+(4-r)²) => x = 5 - 12/7 - sqrt((12/7)²+(16/7)²) => x = 23/7 - sqrt(12²+16²)/7 (12²+16² = 20² still from homothetic 3-4-5 triangle) => x = 23/7 - sqrt(20²)/7 => x = 23/7 - 20/7 = 3/7 (no quadratic equation to solve)

r/(4-r)=(3-r)/r》r=12/7 Potencia de A respecto a la circunferencia =X[(24/7)+X] =[3-(12/7)]^2》X=3/7 Gracias y saludos.

You could just used trig find the angle and r is easily calculated. Great job anyway for that long calculation👍👍

As BC = 4 and CA = 3, ∆BCA is a 3-4-5 Pythagorean triple triangle and AB = 5. Let O be the center point of the semicircle (at the midpoint of PQ). As AP = x and OP = r, AO = x+r. Let M and N be the points of tangency between semicircle O and BC and CA respectively. Draw OM and ON. As radii, OQ = OM = ON = OP = r. First method: As BC and CA are tangent to semicircle O, ∠OMC = ∠CNO = 90°. As ∠MCN = 90° as well, then ∠NOM = 90° and OMCN is a square with side length r. As BC = 4 and CA = 3, BM = 4-r and NA = 3-r. As ∠BMO = ∠BCA = 90° and ∠B is common, ∆BMO and ∆BCA are similar. OB/BM = AB/BC OB/4-r = 5/4 OB = (4-r)5/4 = 5 - 5r/4 AB = 5 AO + OB = 5 (x+r) + (5-5r/4) = 5 x + r = 5r/4 x = 5r/4 - r = r/4 Triangle ∆BMO: OM² + BM² = OB² r² + (4-r)² = (5-5r/4)² = ((20-5r)/4)² r² + 16 - 8r + r² = (400-200r+25r²)/16 32r² - 128r + 256 = 25r² - 200r + 400 7r² + 72r - 144 = 0 7r² + 84r - 12r - 144 = 0 7r(r+12) - 12(r+12) = 0 (r+12)(7r-12) = 0 r = -12 ❌ | r = 12/7 x = r/4 = (12/7)/4 = 3/7 Second method: Draw OC. This creates two triangles, ∆OBC and ∆OCA. As BC and CA are tangent to semicircle O at M and N respectively, OM is perpendicular to BC and ON is perpendicular to CA. [ABC] = [OBC] + [OCA] bh/2 = bh/2 + bh/2 4(3)/2 = 4r/2 + 3r/2 12 = 7r r = 12/7 As OM is perpendicular to BC, ON is perpendicular to CA, and BC is perpendicular to CA, CN = OM = r. As CA = 3, NA = 3-r = 3-(12/7) = 9/7. As ON = 12/7, ∆ONA is a 3/7:1 ratio 3-4-5 Pythagorean triple triangle and AO = 5(3/7) = 15/7. AO = x + r 15/7 = x + 12/7 x = 15/7 - 12/7 = 3/7

You used very long claculations To find r, after proving AEO is similar to ODB AE / OE = OD / BD (3 - r)/r = r/(4 - r) (3 - r) (4 - r) = r^2 12 - 7r + r^2 = r^2 12 = 7r r = 12/7 Now apply Pythagoras in AEO (3-r)^2 + r^2 = (r+x)^2 x = Root((3-r)^2 + r^2) - r Now input r = 12/7 and calculate for x x = 3/7 Alternatively, r / (r+x) = 4/5 (because we have 3-4-5 right triangle) 5r = 4r + 4x r = 4x x = r/4 x = 12/(7 x 4) x = 3/7

It's well triangle OBD is similar to his counterpart OEA also ABC OD is Radius center of semicircle AC=3, r=AE=CE, Square ⬛️ =r*r=r² BC=4, BD+DC=4 , BD+r=4, BD=4-r Theorem (4-r)²+(r)²=16-8r+r²+r² =(BO)² AC=3, AE+EC=3, AE+r=3, 3-r=AE r/(4-r)=(3-r)/r, 12=6r-2r²+4r²+8r-2r² 12=7r, r=12/7 OP=r, QO=r QP=2r=2*12/7=24/7=QP Theorem Phythagorean(degree 90) in triangle BOD (12/7)²(4-12/7)²=(B0)² =(BO)²=8.1632653061 BO=sqrt(8.1632653061) In traingle AOE (r)²+(3-r)²=(r+x)² r²+9-6r+r²=r²+2rx+x² (3-r)²=x(2r+x) x²+24/7x=9-72/7+144/49 7x²+24=(441-504+144)/7 49x²+168=441-504+144 49x²+168=81 (7x)²+(13-1)(13+1)=(9)² x=-b±sqrt(b²-4ac)/2a)=x x=3/7

O needs to be on the 45 deg diagonal through C, and belong also to the line AB. So it's coordinates are (12/7) (-1,1), where C is the origin. The distance of O to A is therefore (15/7). The distance x is therefore 15/7 - 12/7 (the radius of the circle) = 3/7

@RAG981

5 ай бұрын

You mean dist O to A is 15/7, so x is 3/7. Very clever method.

@prbprb2

5 ай бұрын

@@RAG981 You are correct. I tidied things up. Thank you.

Beautiful solution.

5! Satz des Phytagoras: Die Summe der Quadratflächen über den kurzen Seiten eines techtwinkligen Dreiecks, entspricht der Quadratfläche der längsten Seite. (BEWUSST ohne die Verwendung der Fachbegriffe formuliert!) Da die Kantenlänge eines Quadrats die Wurzel aus der Fläche ist: 3×3=9 und 4×4=16 9 +16=25. 5×5=25, also 5. 😂😂😂

With Thales theorem BD/BC=OD/AC (4-R)/4=R/3 we obtain R=12/7 Another Thales theorem AO/AB=OE/BC (x+R)/5=R/4 x=R/4 x=12/4 7=3/7 So easy

r=(A*B)/(A+B) and X=A/(A+B) - simple proportions

The sum of the area of ​​triangle OBD + the area of ​​square OECD + the area of ​​triangle OAE must be equal to the area of ​​triangle ABC. Means a(4-a) + a(3-a) + a^2 =12 This means: a^2 - 7a + 12 = 0 This means: a = 3 or a = 4 While a < 3. This problem is incorrect.

Who says it's a semicircle?

Draw a perpendicular line from the point of tangency from both bases to the hypotenuse, forming a new triangle. The sides of these triangles are 3 - radius, radius, and a hypotenuse of sides 5/4 r since the new triangle is similar to the original 3-4-5 right triangle Hence 5/4 r = (3-r)^2 + r^2 25/16 r^2 = 9 + r^2 - 6r + r^2 25/16 r^2 = 9 + 2r^2 - 6 r 25 r ^2 = 144 + 32 r^2 - 96 r 0 = 7 r^2 - 96 r + 144 r = 1.71429 Hence, the length from the center of the semi-circle to the end of the triangle is 5/4 * 1.71429 or 2.1428625 Since r = 1.71429, then x = 2.1428625 - 1.71429 = 0.4285725

Just use the triangle 3,4,5 You can find it easy. 3.4.5 it's very important and handy learn it

Then you want to complicate what is simple. Triangle ABC is similar to AEO. (3-R)/3 = R/4 = (X+R)/5

The way you this solve problem is way too tedious than it needs be. Just use similar triangles on AOE and OBD to solve for r. Then plug in the value of r in the eqn (r+x)^2=r^2+(3-r)^2 to get x directly. No need to invoke the quadratic formula.

Let's reason about the symmetry of the image in relation to the vertical. The equation of the circle is (x-r)^2+(y-r)^2=r^2. the center is the intersection of the lines: y=x and y=ax+b (noted ∆) with 3=a(0)+b and 0=4a+b i.e.: b=3 and a=-3/4 d 'where y=(-3/4)x+3. For y=x, we will have: (4/4)x=3-(3/4)x or (7/4)x=3. So r=12/7=24/14. If the center is noted C and A(0,3) on ∆, ||AC||=15/7 let D be the point of intersection of the circle and ∆ in the segment AC. ||AC||=||AD||+||DC||=x+r. 15/7=x+12/7, x=3/12. Why ||AC||=15/7 ? ||AC||^2 = r^2+(3-r)^2 = 9-6r+2r^2 = 2(144/49) + (9*49)/49 - (6*12*7)/49 = (288+441-504)/49=(729-504)/49=225/49=(15/7)^2 and then ||AC||=15/7. Other solution : r=12/7 and according to the diagram, ABC and AOE are two identical triangles. Consequently: OE/BC=AO/AB, i.e. r/4=(r+x)/5. So, 5r/20=4r/20+4x/20. So r/20=4x/20, x=r/4=(12/7)/4=3/7.

Check Video Time > 11:00

Con semejanza de triángulos se puede resolver mucho más rápido.

the problem include semicircle? any where?

Your problem not explicit enough. You are assuming that the center of the circle is on the line of ab. The problem doesn’t say that. Therefore the center of the circle is not necessarily be on the hypotenuse also reach the answer thru sin and co-sin is much easier

@MathBooster

5 ай бұрын

I said in the beginning of the video that it is semicircle. So, centre will be on the line AB.

3/(3-r)=4/r=5/r+x

It is much easy to solve... but you went to calculate everything to find your solution. Only the lower triangle solve it.

@j.kl8903

4 ай бұрын

pleasu shut up you are a kid behind the screan who has nothing in his life and talk shit you are not able to solve this so be quit

Откуда следует, что центр окружности на гипотенузе? Кстати она равна пяти.

Устная задача,из подобия тр след (3-R)/3=R/4 ,R=12/7, x=3/7

@Fa-Diez-Major

4 ай бұрын

Еле дошло, (3-R)/R=3/4)))))

It’s also easy by using trignomatry method

Hello, how can we be sure that the center of the circle is on [AB] ?

@rick57hart

5 ай бұрын

Because it is a semicircle.

@gandelve

5 ай бұрын

it is a semicircle

@hoochygucci9432

5 ай бұрын

@@gandelve How do you know it's a semicircle?

@laurentdegara4144

5 ай бұрын

@@rick57hart but it's not written anywhere

@laurentdegara4144

5 ай бұрын

@@gandelve It's not written anywhere.

Complicou demais.

Presh talwaker from mind your decisions made a formula for these types of problems

There is a much easier way to find x by using similar triangles.

Через подобие треугольников будет быстрее и легче!

Similar triangles it is!!

Με ομοιότητα βγαίνει αμμεσα.

3/7 ans

I remember Very in portuguese .

It's not a semi circle.

very lengthy Process....Just use similar triangle method and solve it then it is very easy to simplify..🤬🤣

Thales Theorem it'll be easier.

Чешем левой ногой правое ухо😂😅

❤👍👋

5

This shape does not exist!!! The center of the circle in a triangle like this with sides 4, 3, has hypotenuse =5. The circle tangent to the sides does NOT have a center on the hypotenuse!!! Don't post fake problems! mercy!!! It is impossible that tο be an Olympiad problem!!! Prove it, tell us the year! I'm waiting

Phương Pháp giải hay, nhưng nói tiếng nước ngoài khó hiểu

用这种解法会被人嘲笑一年。

Too long method

aaaarrrgghhhhhh just use a calculatorrr :))))

Дуже погане та нудне навчання блогерами Канади , яке не стимулює учнів : проста задача , яка вирішується двума формулами з подібності трикутників та відношенням квадрата дотичної до сікущої , проведених с однієї точки . OD/BD=AC/DC , r/4-r=3/4 , 4r=12-3r , r=12/7 . AE*2=AP х AQ , (3 - r)*2 = X x (X+2r) (3 - 12/7)*2=Х х (Х + 24/7) , (9/7)*2 = X*2 + 24/7X , X*2 + 24/7X - 81/49 , Х= -12/7 + \/ 144/49+81/49 = -12/7 + \/225/49 = -12/7 + 15/7 = 3/7. Другий корінь негативний , тому його вирішення не приводиться .

note 3/7 = 0.4285725

BO+OA=AB???? Why??????😂😂😂😂😂

0.4285725

5