Find the radius | Nice geometry problem

Find the radius | Nice geometry problem | Math Olympiad

Рет қаралды 8,851

GeometryDose

3 ай бұрын

#geometry
#maths
#mathematics
#triangle
#matholympiadquestion
#rightangletriangle
#circle
#parallel

Пікірлер: 37

@JobBouwman Ай бұрын

Flip a horizontally mirrored copy of this figure onto itself, to see that the diameter equals sqrt(8^2+1^1).

@bpark10001 15 күн бұрын

There is simpler way! Construct chord parallel to BD intersecting AC 2 below point A. This is symmetrical about O. AC is cut into 3 pieces: upper one = 2, mid one = 4, lower one = 2. Center 4 length is symmetrical about O, putting O 2 units above BD. Construct line through center parallel to AC to meet BD at P'. PP' = 1/2 by symmetry. Right triangle BP'O is formed, leg lengths 7/2, 2, hypotenuse = R. R² = (7/2)² + 2². R = √65/2.

@marcgriselhubert3915 3 ай бұрын

We use an adapted orthonormal. P(0;0) A(0;6) B(-3;0) C(0;-2) D(4;0) The equation of the circle is x^2 + y^2 +a.x + b.y +c = 0 with a,b, c unknown. A is on the circle, so 9 -3.a +c = 0; B is on the circle, so 16 +4.a +c =0; C is on the circle, so 36 +6.b +c = 0; D is on the circle, so 4 -2.b +c = 0 The system is very easy to resolve and gives a = -1; b = -4; c = -12, so the equation of the circle is x^2 + y^2 -x -4.y -12 = 0 This equation is also (x -(1/2))^2 + (y -2)^2 = 12 +4 + (1/4) = 65/4. So the radius is sqrt(65)/0 and O(1/2; 2).

@MrPaulc222 2 ай бұрын

I seem to be something of an outlier with the way I do these: Either line may be shifyed, but I will opt to move the horizontal chord upwards, purely because the vertical is an even-numbered length and it makes the calculation slightly easier. Move the horizontal chord up to pass over the diameter.This gives chord sections of 4 and 4 for the vertical and (4+x) and (3+x) for the horizontal. x is the distance between the ends of the chord (after moving it) and the circumference. It is the same distance on both sides so can have the same variable name. 4*4=(x+4)(x+3) 16 = x^2 + 7x + 12 (we need to find the value of 2x and add it to the chord length 7. Quadratic formula: x^2 + 7x - 4 = 0 (-7+or-sqrt(49-4*1*-4))/2 = x (-7+or-sqrt(65))/2 = x. Actually, I can dispense with the /2, because I want 2x. 2x = sqrt(65) - 7, so the diameter is 7 + sqrt(65) - 7, which simplifies to sqrt(65). r = sqrt(65)/2 In decimal this is 4.031 (rounded).

@geoninja8971 27 күн бұрын

three of the points form a triangle with apices on the circle - its easy to work out the circumcentre with no fuss...

@ivorauh341 Ай бұрын

why so difficult? as cords intersect exactly at the middle to the center point only one pythagoras is needed: 2**2 +3,5**2=r **2 direct result r=sqrt 65 divided by 2

@55dembel 29 күн бұрын

Exactly! Didn't even need to draw anything, just looked at the picture. for a minute)

@RAG981 23 күн бұрын

It seems there are lots of easier ways! Mainly just using the perp bisector of a chord passes through the centre, and Pythagoras of course.

@pk2712 20 күн бұрын

You are definitely correct ; but , doing it the easy way would only result in a video of about one minute duration .

@johnbrennan3372 Ай бұрын

Through O draw diameter perpendicular to AC. Then (r-1/2)(r+1/2)=(4)(4) so r^2-1/4=16. So r^2=65/4 giving r=sqroot65 /2

@larswilms8275 Ай бұрын

Nice solution. Next time though please explain how you got the 1/2 added and subtracted from the radius from. You now bring it as a big surprise. It comes from the distance from the O to the line AC. And can be calculated using line BD (BD = BP + DP = 3+4 = 7) The midpoint of BD is at the same distance from P as O from line AC, The midpoint is 7/2 from B, therefor the distance from the midpoint is 1/2

@johnbrennan3372 Ай бұрын

@@larswilms8275 Thanks for your reply. You worked out yourself very nicely how the 1/2 came about.That’s exactly how I worked it out myself.

@brettgbarnes Ай бұрын

The coordinates of the four points are obvious. A: ( -0.5 , 4 ) B: ( -3.5 , -2 ) C: ( -0.5 , -4 ) D: ( 3.5 , -2 ) r² = x² + y² Using Point A: r² = (-1/2)² + 4² r² = 1/4 + 16 r² = 65/4 r = (√65)/2

@calvinmasters6159 Ай бұрын

Fun. Took the image, rotated 180deg and superimposed on itself. Produced a 1 x 4 rectangle centered symmetrically around circle center. r^2 = 16^2 + 0.5^2

@palmcrustaustralia9849 Ай бұрын

It's much simpler than that. Draw OM ⟂ AC (M∈AC), so AM=CM=(6+2)/2 = 4. Then draw ON ⟂ BD (N ∈ BD), and AN = DN = (3+4)/2 = 7/2. Therefore PN= PD-DN = 4-7/2=1/2 and OM = PN = 1/2. From triangle OMA, R = OA = √(AM²+OM²) = √(4²+(1/2)² = √(4²+(1/2)²)=√(16+1/4) = √65 / 2. BTW, condition PC=2 is redundant. It isn't used here, but anyway it can evaluated, given that BP x PD = PC x PA.

@KarlFrei 3 ай бұрын

Very nice problem! I think it can be solved in a somewhat simpler way though 🙂 Draw a line EF that is parallel to BD in the top half of the circle at distance 2 from A. By symmetry, the center of the rectangle EFDB is O. The vertical line through O intersects BD at point Q. The right-angled triangle BOQ has sides 3.5 (BQ) and 2 (OQ, because the distance from BD to EF is 4), so the radius of the circle, which is the length of BO, is the square root of 16.25.

@GeometryDose 3 ай бұрын

I agree there are indeed simpler ways to solve this for the radius. The main purpose I made this video was to show how to derive the formula (2R)^2= AP^2 + BP^2 + CP^2 + DP^2. But I can't put this whole formula on the title of the video because fewer people will click the video that way so I hope you understand.

@Emil_Avg 3 ай бұрын

If you want a fast solution you could for example calculate the area of triangle ABC and then use the formula which connects this area with the radius of the circumcircle.

@hanswust6972 Ай бұрын

Man, more than solving a problem you proved a theorem!

@GeometryDose Ай бұрын

Actually the main purpose I made this video was to prove this theorem. Finding the radius was just a little bonus. Glad you noticed it :).

@PrithwirajSen-nj6qq 19 күн бұрын

Just draw perpendicular on each of the chords from the centre. Centre = C Perpendicular on chord (7 units) =CP Perpendicular on chord (8 units)=CM Intersecting point of chords =N CPNM a rectangle. CP=MN MN=1/2(6+2)-2=2 CP=2 🔺 CPR (R is the point that intersects circumference and the 7 unit chord (right hand side) Hence CR =√[(7/2)^2+2^2] Comment please

@GeometryDose 19 күн бұрын

Nice

@arcticantic1768 3 ай бұрын

answer is square root of 17. Draw vertical line through center O, and horizontal line through the center O. Connect O with A (it will be a radius, R). You will have right triangle A,O,crosspoint of AC and horizontal line. You will have triangle with sides 4,1,R. square of R=16+1=17.

@Masterclass_Geometry 2 ай бұрын

great👌

@albaldin 2 ай бұрын

No. The sides of the triangle are 4, 1/2, R

@Grizzly01-vr4pn Ай бұрын

@@Masterclass_Geometry No, not great. 'Wrong' is the word you're grasping for. I'd have hoped for better than that from someone who runs their own geometry channel ☹

@Grizzly01-vr4pn Ай бұрын

@@albaldin or if you use the same procedure, but 'horizontally' rather than 'vertically', you get a right triangle with legs 3½ and 2, which gives the same hypotenuse (= radius) of (√65)/2.

@albaldin Ай бұрын

@@Grizzly01-vr4pn Square of ((3/2) + 2) is 25/4 (root 5/2). It's 7/2 and 2