Romania Math Olympiad Problem | Very Nice Geometry Challenge | 2 Methods

Romania Math Olympiad Problem | Very Nice Geometry Challenge | 2 Methods
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[@oscarcastaneda5310]

I found it fun to demonstrate that for every quarter circle of radius "R", the wanted distance is 1.2R : )

[@ashugarg5475]

Once you have calculated theta, calculate x by using cosine rule in triangle AOC, angle OAC =2*theta. And OA=OC=6. X comes to be 7.2

[@user-kb1vm8qk7x]

Let us denote the point of tangency between the segment and the semicircle as D1 AD1 = AO. Let us extend the segment AC with end D. Using the tangent and secant formula, we find BD. Draw the radius OC and get two isosceles triangles. Angle AOC and COB give 90 degrees, knowing this we find that angle OCA + OCB = 135. Angle BCD = 180 - 135 = 45. Draw the radius of the semicircle and label the end D1. Angle O1C1D=90. We get a Pythagorean triple. From triangle O1C1D we obtain the sine and cosine of angle D. Next we use the sine theorem to find CB, and then the cosine theorem to find CD. Well, then we count and check using the Pythagorean theorem :) Точку торкання відрізка та півкола позначимо D1 AD1 = AO. Продовжимо відрізок AC з кінцем D. За формулою дотичної та січної знаходимо BD. Проводимо радіус OC і отримуємо два рівнобедрених трикутники. Кут AOC і COB дають 90 градусів, знаючи це ми бачимо, що кут OCA + OCB=135. Кут BCD = 180 - 135 = 45. Проводимо радіус півкола і позначимо кінець D1. Кут O1C1D=90. Отримуємо піфагорову трійку. З трикутника О1С1D отримуємо синус і косинус кута D. Далі використовуємо теорему синусів для знаходження CB, а далі теорему косінусів для знаходження CD. Ну а далі рахуємо та перевіряємо за теоремою Піфагора :)

[@robertlynch7520]

I'm glad I didn't skip ahead. Solved this another way, and it was a good one. My strategy was • draw a line from upper corner to center of the semi-○ • identify lengths, and use Pythagorean method • describe the 'formula for a ○' for the quarter-○. • and the given line which is tangent to the semi-○. • let them be equal formulæ, solving for the intersection. • use that to determine Δ𝒙 and Δ𝒚 • which in turn by Pythagoras, determines length of the segment ("goal!") So… [1.1]  (3 + 𝒋)² = 3² ⊕ 6² … for the semi-○-to-corner segment [1.2]  3 + 𝒋 = √(45) Now with that, find the length of the given segment from upper corner to intersection with semi-○ Note that it has the same hypotenuse (√(45)) and the same little leg (3) so it must be [2.1]  𝒌 = 6 Not very exciting. Next however, need the slope of that given segment. One way would be to figure the angle θ at the upper-left point of the 6:3 left △ … which is [3.1]  θ = arctan( 3 ÷ 6 = 0.5 ) [3.2]  θ = 26.565° Clearly, just need to double θ then find its tan() function: [4.1]  tan( 2θ ) = 1.3333… How does this help? It says that the ratio of base to height is 1.333 or ⁴⁄₃. But the slope is the inversion of that [5.1]  g(𝒙) = -¾𝒙 + 6 The [⊕6] satisfies the [0, 6] condition. With that formula-of-a-line, we now need to intersect it mathematically with the formula of the quarter-○ [6.1]  h(𝒙) = √(6² - 𝒙²) … and set this to the line function [6.2]  -¾𝒙 ⊕ 6 = √(6² - 𝒙²) … square both sides [6.3]  ⁹⁄₁₆𝒙² - 9𝒙 + 36 = 36 - 𝒙² … cancel the 36's and move bits about [6.4]  𝒙² + ⁹⁄₁₆𝒙² - 9𝒙 = 0 … divide by 𝒙 (losing the quadratic making it simpler) [6.5]  ²⁵⁄₁₆𝒙 - 9 = 0 … and solve for 𝒙 [6.6]  𝒙 = 5.76 Now figure out the (𝒚) of this 𝒙 using either equation (the linear one is easier) [7.1]  𝒚 = -¾ × 5.76 ⊕ 6 [7.2]  𝒚 = 1.68 And then the length of the sought after line segment is by Pythagoras [8.1]  Δ𝒛 = √(Δ𝒙² + Δ𝒚²) [8.2]  Δ𝒛 = √(5.76² + (6 - 1.68 = 4.32)²) [8.3]  Δ𝒛 = 7.2 And that's the solution to the problem requested. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

[@User-jr7vf]

I did the first method (without looking at the solution), which is the quickest one. However the second method and yours are more interesting, since they require to think outside the box.

[@brettgbarnes]

If you add to the diagram a radius from O through D to the circumference you can use Pythagoras Theorem, similar triangles scaling factors, and intersecting chords theorem to solve for AC.

[@richardleveson6467]

Thank you - very nice!

[@ryanchin5232]

3rd method: using the interception point/tangency of y - 6 = m x and (x-3)^2 + y^2 = 3^2 solve the system. Because there is only one interception, so the discriminant for quadratic formula b^2 - 4ac=0. Then the slope m can be found m = - 3/4. Next, solve the system of y = - 3/4 x + 6 and x^2 + y^2 = 6^2. The interception point is (144/25, 42/25). Finally, find the distance between two points, (0, 6) and (144/25, 42/25). That distance is 7.2.

[@franciscook5819]

Take the diagram at 9:56. tan θ = 3/6 = 1/2 tan 2θ = 2*(1/2)/(1-(1/2)²)=1/(3/4)=4/3 (formula) let the angle at E be α tan α = cot 2θ = 3/4 treat the problem as a locus question: line y=mx+c (EA) meets circle AB (x²+y²=6²=36) line: x=0, y=6 => c=6 ; gradient = -3/4 (calculated above) y=-(3/4)x+6 y²=(9/16)x²-9x+36 sub into circle eqn x²+(9/16)x²-9x+36=36 (note the 36 cancels) x((1+9/16)x-9)=0 x(x25/16-9)=0 so x=0 or x=144/25 with (using y=mx+c) y=6, y=42/25 respectively AC = sqrt((y0-y1)²+(x0-x1)²) =sqrt((6-42/25)²+(0-144/25)²) =(1/25)*sqrt(108²+144²) =(1/25)*sqrt((3*36)²+(4*36)²) =(36/25)sqrt(9+16) =36/5 =7.2

[@soli9mana-soli4953]

Sir, in your first method once found cos 2theta = 3/5 you can trace the radius perpendicular to line AC that splits AC in two equal parts, then x/2 = 6*cos 2theta x = 12 * 3/5 = 36/5

[@rabotaakk-nw9nm]

👍🥰

[@soli9mana-soli4953]

@@rabotaakk-nw9nm❤

[@RobertHering-tq7bn]

Hi, following my solution by using coordinates, line equations and more. I named the center of the semicircle as D and the point where AC touches the semicircle as E. In addition I renamed "x", that we are looking for, to "a", because I'm using x and y as variable names for the functions that we deal with. Coordinates of important points... O(0,0), A(0,6), D(3,0), E(Ex,Ey), C(Cx,Cy), B(6,0) The lines thru A and C and then thru D and E stand perpendicular to each other. Therefore the product of their slopes is (-1). Line thru A and C... (y-6)/(x-0) = m1 Line thru D and E... (y-0)/(x-3) = m2 We know... m1*m2 = -1 and the lines are crossing each other in E... ((Ey-6)*Ey)/(Ex*(Ex-3)) = -1 Ey² - 6Ey = 3Ex - Ex² (1) Ex² + Ey² = 3Ex + 6Ey Looking to DE especially, we see by Pythagoras... 3² = (Ex-3)² + (Ey-0)² 9 = Ex² - 6Ex + 9 + Ey² (2) Ex² + Ey² - 6Ex = 0 Subtracting (2) from (1)... 6Ex = 3Ex + 6Ey 3Ex = 6Ey (3) Ex = 2Ey Using (3) in (2)... (2Ey)² + Ey² - 6(2Ey) = 0 5Ey² - 12Ey = 0 ==> 5Ey - 12 = 0 as Ey 0 So... Ey = 12/5 and Ex = 2Ey = 24/5 Now looking to point C on line thru A and E and on the quartercircle from A to B, starting with the line... (y-6)/(x-0) = ((12/5)-6)/((24/5)-0) (y-6)/x = (12-30)/24 = (-18)/24 = -(3/4) (4) y = -(3/4)x + 6 On the quartercircle we have... (5) x² + y² = 6² = 36 With (4) and (5) we get... Cx² + (-(3/4)Ex+6)² = Ex² + (9/16)Ex² - 2*6*(3/4)Ex + 36 = 36 (25/16)Ex² - 9Ex = 0 ==> (25/16)Ex = 9 as Ex0 So... Ex = 144/25 Now... Ey = -(3/4)*(144/25) + 6 = 6 - 108/25 = (150-108)/25 = 42/25 Finally we get with Pythagoras... a² = (6-42/25)² + ((144/25)-0)² = (108/25)² + (144/25)² = (11664+20736)/25² = 32400/25² = 180²/25² So... a = 180/25 = 36/5 = 7.2 and we are done...

[@jimlocke9320]

We note that tan(Θ) = OP/OA = 3/6 = 1/2. Applying the tangent double angle formula, tan(2Θ) = (2tan(Θ))/(1 - tan²(Θ)) = (2(1/2))/(1 - (1/2)²) = 1/(1 - 1/4) = 4/3. At 9:55, we note that tan(2Θ) = OE/OA, so 4/3 = OE/6 and OE = 8. BE = OE - OB = 8 - 6 = 2. We observe that right ΔOAE has sides 6 and 8, so it is a Pythagorean triple 3 - 4 - 5 right triangle scaled up by a factor of 2 and hypotenuse AE = 10. So, CE= AE - AC = 10 - x. We have found enough lengths to skip ahead to 16:25 and complete the calculation of X.

[@abcdsah]

Draw a perpendicular from O to AC. AC = 2(6-x/2) Now, 6² = (x/2)²+(2(6-x/2))² Solving, x = 36/5

[@zdrastvutye]

there are 2 different ways to solve this- the direct product or the thales circle: 10 print "mathbooster-romania math olympiad-very nice geometry":nu=35 20 dim x(2),y(2):l1=6:r1=l1:r2=l1/2:sw=l1/10:xd=sw:goto 50 30 yd=sqr(r2^2-(xd-r2)^2):dgu1=(r2-xd)*xd/l1^2:dgu2=(l1-yd)*yd/l1^2 40 dg=dgu1+dgu2:return 50 gosub 30 60 dg1=dg:xd1=xd:xd=xd+sw:xd2=xd:gosub 30:if dg1*dg>0 then 60 70 xd=(xd1+xd2)/2:gosub 30:if dg1*dg>0 then xd1=xd else xd2=xd 80 if abs(dg)>1E-10 then 70 print xd,yd 90 lad=sqr(xd^2+(yd-l1)^2):l=sw:dx=xd/lad:dy=(yd-l1)/lad :goto 120 100 dxl=l*dx:dyl=l*dy:xc=dxl:yc=l1+dyl:dgu1=xc^2/l1^2:dgu2=yc^2/l1^2 110 dg=dgu1+dgu2-1:return 120 gosub 100 130 dg1=dg:lu1=l:l=l+sw:lu2=l:gosub 100:if dg1*dg>0 then 130 140 l=(lu1+lu2)/2:gosub 100:if dg1*dg>0 then lu1=l else lu2=l 150 if abs(dg)>1E-10 then 140 160 print "die gesuchte entfernung=";l 170 mass=850/l1:goto 190 180 xbu=x*mass:ybu=y*mass:return 190 xmu=0:ymu=0:ru=l1:we=pi/2:gosub 200:goto 240 200 x=ru:x=x+xmu:y=ymu:gosub 180:xba=xbu:yba=ybu:for a=1 to nu:wa=we*a/nu 210 x=ru*cos(wa):x=x+xmu:y=ru*sin(wa):y=y+ymu:gosub 180:xbn=xbu:ybn=ybu:goto 230 220 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 230 gosub 220:next a:return 240 ru=l1/2:xmu=ru:we=pi:gosub 200:x(0)=0:y(0)=l1:x(1)=xd:y(1)=yd:x(2)=xc:y(2)=yc 250 x=x(0):y=y(0):gosub 180:xba=xbu:yba=ybu:for a=1 to 2:x=x(a):y=y(a):gosub 180 260 xbn=xbu:ybn=ybu:gcola:gosub 220:next a 270 move 0,50:gcol8:print "run in bbc basic sdl and hit ctrl tab to copy from the results window" mathbooster-romania math olympiad-very nice geometry 4.8 2.4 die gesuchte entfernung=7.2 run in bbc basic sdl and hit ctrl tab to copy from the results window >

[@batmunkhenkhbaatar9061]

my answer is 7.5

[@vcvartak7111]

Nice problem

[@someonespadre]

This is a pretty simple using trigonometry.

[@MarieAnne.]

I used 2 different methods. The first method I used was a combination of the 2 methods shown. Find tan(θ) = 1/2 and cos(2θ) = 3/5 as shown in method 1 of video. Then complete the circle as shown in method 2 of video. But instead of extending BO to point M, we extend AO down to point N, then we draw line CN We now have △ACN, with AC = x, AN = 12 (diameter of circle) Since AN is a diameter of circle, then ∠ACN = 90°, so in right △ACN we get: cos 2θ = AC/AN 3/5 = x/12 x = 36/5 = 7.2 The second method I used was coordinate geometry, with point O at origin: O = (0,0), A = (0,6), B = (6,0) Let P = center of semi-circle → P = (3,0) Line through A and C has y-intercept = 6. So it has equation: y = mx + 6 Point P is below this line, so it is in the region y > mx + 6 → mx − y + 6 > 0 Since this line is tangent to semi-circle centered at P, then distance from P to line mx − y + 6 = 0 is 3 (radius) Using point-line distance formula, we get: d = |mx₀ − y₀ + 6| / √(m²+(-1)²) = 3 where P = (x₀, y₀) = (3, 0) Since mx − y + 6 > 0 for point P, we get: (m(3) − 0 + 6) / √(m²+1) = 3 3m + 6 = 3√(m²+1) m + 2 = √(m²+1) m² + 4m + 4 = m² + 1 4m = −3 m = −3/4 Line through A(0,6) and C has equation y = −(3/4)x + 6 Both A and C are located on circle x² + y² = 6² Substituting, we get: x² + (−(3/4)x + 6)² = 36 x² + (9/16 x² − 9x + 36) = 36 25/16 x² - 9x = 0 25/16 x (x - 144/25) = 0 x = 0 or x = 144/25 = 5.76 Point A: x = 0, y = −(3/4)(0) + 6 = 6 Point C: x = 5.76, y = −(3/4)(5.76) + 6 = 1.68 AC² = (0−5.76)² + (6−1.68)² = 51.84 AC = 7.2

[@rocksalt636]

If you ever got a geometry problem just use triangles.

[@sourabhbhattacharya2299]

AD = 6 and OC also be = 6, how is it possible?

[@ExpressStaveNotation]

I did it in my head, 36/5.

[@ExpressStaveNotation]

Find tan of the angle by noting that tan of half the angle is 1/2. (=4/3) This gives a 3,4,5 right triangle. So the right triangle formed by bisecting arc X is also (3,4,5). So X/2 is 3/5 of 6.