USA Math Olympiad | A Very Nice Geometry Problem

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USA Math Olympiad | A Very Nice Geometry Problem
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Пікірлер: 28

@jimlocke9320 3 ай бұрын

Alternative method of computing x (approx. 4:50 into the video): Construct OC and OF. Note that OC bisects

@TheAndreArtus Ай бұрын

I love these problems, I always attempt them before watching the rest of the video. And I love this channel: it's one of few channels where reading through the comments is actually enlightening and not mostly depressing.

@Mediterranean81 3 ай бұрын

Fact : Every circle inscribed in a right (3;4;5) triangle has a radius of 1

@quigonkenny 3 ай бұрын

Not only that, but the ratio remains true for any similar triangles. So R = 2 for a 6-8-10 right triangle, and R = 3 for a 9-12-15 right triangle, etc.

@Mediterranean81 3 ай бұрын

@@quigonkenny yes

@RahulKumar-id5cq 3 ай бұрын

I think two cases arises 1.When. JD is tangent line and touches the circle at point K(Let) above J. 2.line JD is intersecting and intersect circle at J and J'(let above J).

@RahulKumar-id5cq 3 ай бұрын

As angle BCA=37° (from triangle HEC) Now in ∆ABC let AC=x and angle BAC=37°; So in-Circle radius will be x(Sin37°+Cos37°-1)/2=1/5 x; Now BC=BD+DE+EC =2R+3+4=2/5 x+7; but BC =xCos37(from ∆ABC) =4x/5; so x=35/2(Equatting both BC) R=1/5 x=7/2.

@michaeldoerr5810 3 ай бұрын

This is a nice problem in that it makes use of two circle theorems and the extension of the second circle theorem is what justifies the angle bisector. Along the way putting the first circle theorem in practice shows eventually that this is NOT a 3-4-5 triangle and because of that you have to use process of elilimination in order to evaluate how much is the radius of the insribed circle. And what follows is using delta and the Pythagoras Theorem in order find the radius. I almost forgot to mention that you have to keep track of r.

@michaeldoerr5810 3 ай бұрын

And finding R actually starts with setting two sides equal as per the angle bistctor extension relative to the first circle theorem.

@michaeldoerr5810 3 ай бұрын

I also appreciate the calculator free method of using the quadratic eqn. Looks like I will have to use this problem and time myself at the exact amt of time shown.

@RahulKumar-id5cq 3 ай бұрын

For Case 1 ; R=7/2 Case 2 (but here we have to assume R>3) So,R=(8+√6)/3.

@quigonkenny 3 ай бұрын

Point labels: Small circle center: P Large circle center: O Triangle (ccw from top left): ABC Square (ccw from top left): DEFG Points of tangency of small circle: J (on FG), K (on BC), L (on CA) Points of tangency of large circle: M (on AB), N (on BC), S (on CA) Let R be the radius in the large circle. In circle P, draw radii PJ, PK, and PL. All are length 1, being radii of circle P. As FG is tangent to circle P at J, and GC is tangent to circle P at K, ∠PJF = ∠FKP = 90°. As ∠JFK = 90° as well, ∠KPJ must also equal 90°, and so PJFK is a square with side length 1. As JF = 1 and FG = 3, GJ = 2. LG and GJ are tangents to circle P that intersect at G, so LG = GJ = 2. Let KC = x. As FK = 1, FC = x+1. KC and CL are tangents to circle P that intersect at C, so CL = KC = x, and thus CG = CL+LG = x+2. Triangle ∆GFC: FG² + FC² = CG² 3² + (x+1)² = (x+2)² 9 + x² + 2x + 1 = x² + 4x + 4 10 - 4 = 4x - 2x 2x = 6 x = 3 Therefore FC = 1+3 = 4 and CG = 2+3 = 5, and ∆GFC is a 3-4-5 Pythagorean triple right triangle. Note that as FG and AB are both perpendicular to BC and are thus parallel, CG is collinear with CA, FC is collinear with BC, and ∠C is common, ∆GFC and ∆ABC are similar triangles. Draw ON, OM, OS, and OD. As radii of circle O, all are equal to R. As AB is tangent to circle O at M and BC is tangent to circle O at N, ∠OMB = ∠BNO = 90°. As ∠MBN = 90° as well, ∠NOM must also equal 90°, and so OMBN is a square with side length R. Let NE = y. NC and CS are tangents to circle O that intersect at C, so CS = NC = y+7. Draw OC. As OS = ON, NC = CS, and OC is common, ∆ONC and ∆CSO are congruent triangles, and thus OC is an angle bisector of ∠NCS. As PK = PL, KC = CL, and PC is common, ∆PKC and ∆CLP are congruent and PC is an angle bisector of ∠KCL. Thus OC and PC are collinear, ∠KCP = ∠NCO, and ∆ONC is similar to ∆PKC. ON/NC = PK/KC R/(y+7) = 1/3 y + 7 = 3R y = 3R - 7 R appears to be longer than the square side length (3) from the figure, but it is unclear. If R is ≤ 3, then DE is tangent to circle O and NE = y = R: y = R 3R - 7 = R 2R = 7 R = 7/2 But 7/2 > 3, so R cannot be ≤ 3. Draw DT, where T is the point on ON where DT is perpendicular to ON. As ∠DTN = ∠TNE = ∠NED = ∠EDT = 90°, DTNE is a rectangle with height 3 and width y, therefore DT = NE = y = 3R-7. As TN = 3 and ON = R, OT = R-3. Triangle ∆OTD: OT² + DT² = OD² (R-3)² + (3R-7)² = R² R² - 6R + 9 + 9R² - 42R + 49 = R² 9R² - 48R + 58 = 0 R = -(-48)±√(-48)²-4(9)(58) / 2(9) R = 48/18 ± √(2304-2088)/18 R = 8/3 ± √216/18 R = 8/3 ± 6√6/18 = (8±√6)/3 R = (8-√6)/3 ❌ 3 R = (8+√6)/3 ✓

@sarvajagannadhareddy1238 2 ай бұрын

Dear, please let me know 1. what is BC ? and 2. what is AB length ?

@Grizzly01-vr4pn 2 ай бұрын

Well, as we know, △ABC ⁓ △HEC, so AB = 3R = 8 + √6 and BC = 4R = (32 + 4√6)/3

@sarvajagannadhareddy1238 2 ай бұрын

@@Grizzly01-vr4pn ;; Dear , thank you. GOD BLESS YOU AND YOUR FAMILY

@abdesselemelhassouni281 3 ай бұрын

I don’t understand I just applied the 2 tangents method to the big cercle and get immediately R=3. Then I thought that DJ is not a tangent to the big cercle and that will lead to get a radius of the small cercle not equal to one in this case. But after double check the radius of the small cercle is 1, so no mistake. Anyone see where I’m wrong?

@Grizzly01-vr4pn 2 ай бұрын

DJ is definitely _not_ a tangent to the big circle. That is what 12:50 onwards is all about. EH however, definitely _is_ tangent to the small circle.

@TheAndreArtus Ай бұрын

If you look closely you will see that the circle origin is above a line drawn through JH. If the circle was tangent to a line going through DJ then you would not have to do any of the other calculations, R would be 3 (but it is not).