China Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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- Опубліковано 6 лип 2024
- China Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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15:04 Once you've found PA, you can you can directly find PC using Pythagoras theorem (PBC is a right angled triangle). PC is the diameter as it projects a right angle on the circumference at B. PC /2 is the radius.
Let A(0,0), then B(6,0), C(6,-13), D(0,-18)
O is the centre of the circle, then O(k,-13/2)
OC=OD
(k-6)^2+(13/2)^2=k^2+(23/2)^2
k=-9/2
radius=OC=sqrt((-9/2-6)^2+(13/2)^2)=sqrt(610)/2
Another way to do it is to draw the segment lines from O to D and B which are the radius of the circle. name x the distance from the center O to M. We consider the right triangles ONB and OMD. Observe that the length of the side BN is 13/2 and the length of MD is 18-13/2=23/2. Now use Pythagoras theorem for both triangle to write: R^2=(x+6)^2+(13/2)^2 and R^2=x^2+(23/2)^2. From this point no more geometry, now only algebra. It follows that (x+6)^2+(13/2)^2=x^2+(23/2)^2. This equation simplifies itself into 12x-54=0 or x=9/2. Therefore, R^2=(9/2)^2+(23/2)^2=81/4+529/4=610/4. Finally, R=sqrt(610)/2.
Extending BA to the left side of the circumference and a parallel chord from C also to the left circumference, we see there are 5 units below 13 units in the extended AD chord is 23 units, and the extended chord of AB is 21 units long. Center of the circle is at (21/2, 23/2) = (10.5, 11.5). Now R^2 - (11.5) ^2 + (10.5 - 6) ^2 = 11.5^2 + 4.5^2 = [23^2+9^2]/4 = (529+81)/4 = 610/4 -->
R=sqrt (610) /2.
I extended the 18 line to meet the circle, giving a line length 23 (symmetry). I constructed the perpendicular bisector of the 13 chord (which must pass through the centre of the circle and bisect the line length 23).
If left intersect of perp to line 18 or 23 is "a" and from there to right intersect is "b" then we have two equations in a,b using the two chords length 23 and 13:
a.b=23²/4 and (a+6)(b-6)=13²/4 which can be solved by substituting for b in the second eqn using the first eqn.
Then a+b=2r so r=(a+b)/2
Let D=(0,0), then C=(6,5), B= (6,18)
Let O be the center. Since the horizontal line from passing O bisects BC, we know O=(x, 11.5)
Let M be the midpoint of C and D, we know M=(3, 2.5)
Since OM is perpendicular to CD, we have (x-3,9) inner product with (6, 5)=0
6x-18+45 = 0, hence x=-4.5
Therefore, the radius r=\sqrt{(-4.5)^2+11.5^2}=(1/2)\sqrt{9^2+23^2}=sqrt{610}/2
(x-h)^2+(y-k)^2=r^2 B(x,y) = 0,0 C=0-13 D=-6,-18
then h^2+k^2=r^2 ...... eq1
h^2+-(13-k)^2=r^2 ...........eq2
(-6-h)^2+(-18-k)^2=r62 ......eq3
solve eq1=eq2 k=169/26
solve eq3=eq1 360-12h+h^2-36k+k^2=r^2 suitation k h=[360-(36*19/26)]/12
suitation h k eq1 r=12.349 As same answer
Alternative "Insane" Method
For number-crunching lovers only. If you're at all error prone with long complicated equations... don't even bother. If you like spending at least half an hour doing tons of algebra on a single problem... go for it. It took me three attempts, but it does work.
Use chord bisector radius to create two Pythagoras equations.
r² = (r - x)² + (13/2)²
r² = [r - (x + 6)]² + [18 - (13/2)]²
Have fun!
after 15:00 :Use Pythagorean theorem on triangle PBC
Yes,
ótimo
I think it’s much simpler to use the formula for a circle. (x-a)^2 + (y-b)^2 = r^2 where a and b are the center coordinates of the circle.
The intersection points are (0,0),(6,5)and(6,18) (arbitrary origin on bottom point).
Then solve the three equations generated by the three intersection points. Easy!
If I understand these methods, the first method triangulates to find the radius using the Pythagorean Thrm twice then uses a circle theorem and the second method requires applying chords twice. I thnk that this is the first time chords are diffcult to visualize.
On second thought, I understood both methods!!!
Las alineaciones BA y DA cortan la circunferencia en los puntos F y E respectivamente→ Potencia del punto A respecto a la circunferencia = AB*AF=AD*AE→ 6a=18*(18-13)→ a=15 → FC²=BF²+BC²→ (2r)²=(6+a)²+13²=21²+13²→ r=(√610)/2 =12,34908....
Gracias y un saludo cordial.
18-13=5 18*5=x*6 6x=90 x=15
(15+6)/2=21/2 (18+5)/2=23/2
R²=(21/2-6)²=(23/2)² R²=81/4+529/4=610/4 Radius of the circle = √610/2
Let O be the center of the circle. Draw radius OT, so that OT is perpendicular to BC and intersects it at M. As OT is perpendicular to BC and BC is parallel to AD, OT is also perpendicular to AD, and intersects it at N.
As BC is a chord, and as OT is an intersecting radius that is perpendicular to BC, then OT bisects BC, thus BM = MC = 13/2.
As ∠ANM = 90° and ∠NMB = 90°, then ABMN is a rectangle, and thus NM = AB = 6 and AN = BM = 13/2. As AD = 18, ND = 18-13/2 = 23/2.
Extend AD from A to E so that ED is a chord. OT bisects ED for the same reasons as it does BC, therefore EN = ND.
EN = ND
EA + AN = ND
EA + 13/2 = 23/2
EA = 10/2 = 5
By the intersecting chords theorem, if two chords of a circle intersect, then the products of the lengths of each side of the intersection are equal for each chord. Extend AB from A to F, so that FB is a chord. As A is the intersection point between FB and ED, then FA•AB = EA•AD.
FA•AB = EA•AD
6FA = 5(18) = 90
FA = 90/6 = 15
Method 1:
Draw OB, and draw radius OS such that OS is perpendicular to FB and intersects it at P. As above, OS bisects FB, so FP = PB = (15+6)/2 = 21/2. As PB is perpendicular to BC, PB = OM = 21/2. As BM is perpendicular to FB, OP = BM = 13/2.
Triangle ∆OBM:
BM² + OM² = OB²
(13/2)² + (21/2)² = r²
r² = 169/4 + 441/4 = 610/4
r = √(610/4) = √610/2
Method 2:
If two chords intersect perpendicularly, then the sum of the squares of the divided chord segments equals 4 times the radius squared.
AB² + FA² + AD² + EA² = 4r²
6² + 15² + 18² + 5² = 4r²
4r² = 36 + 225 + 324 + 25 = 610
r² = 610/4
r = √(610/4) = √610/2
R^2=6,5^2+a^2...R^2=(18-6,5)^2+(a-6)^2...a=10,5...R^2=6,5^2+10,5^2=152,5
Second Method until the value of AP=15 is excellent.But then misses the obvious, that Inscribed Angle
(18]^2= 244 (13)^2 =169 (6)^2 =36 {244A+169B+36C}= 419ABC 419ABC/360°=1.59ABC (ABC ➖ 59ABC+1)
At 13:39 PBC is an inscribed rectangle triangle then CP is a diameter; from 14:58 all you need is to use Pythagoras to find |CP|, and then R=|CP|/2: |CP|=√(21²+13²)=√(610), R= √(610)/2.
Glad I'm not the only one who thought of this. As soon as I saw the diagram, I thought of extending BA and using the right triangle.
Why not just use triangles OFM and OBN? We will have r^2=OM^2+MF^2 and r^2=ON^2+NB^2. NB=13/2=6.5, MF=MD=18-6.5=11.5, ON=OM+6. Let OM=x, then r^2=x^2+11.5^2, r^2=(x+6)^2+6.5^2. Subtracting first from second: 0=(x+6)^2+6.5^2-x^2-11.5^2=2*x*6+6^2+6.5^2-11.5^2;x=((11.5-6.5)(11.5+6.5)-36)/12=(5*18-36)/12=54/12=27/6. Then r=sqrt(27^2/6^2+11.5^2)=sqrt(152.5)=12.349.....
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Excelente solução! 🎉
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Muito bom e muito obrigado!
La línea AD pasa por el centro del círculo?
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2nd Method much better thanks
EC in triangle DCB is not completely the height
Yes it is.
سخيف