Poland Math Olympiad | A Very Nice Geometry Problem

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[MarieAnne.]

Let O = center of circle. OA = OP = OQ = OB = radius = x/2
Let ∠AOP = ∠POQ = θ → ∠BOQ = 180 − 2θ
Using law of cosines in △AOP we get:
cos θ = = (x²−18)/x²
Using law of cosines in △BOQ we get:
cos(180−2θ) = = (x²−98)/x²
Now we setup relationship between these angles to get equation in x only
cos 2θ = −cos(180−2θ)
2 cos²θ − 1 = −(x²−98)/x²
2 (x²−18)²/x^4 − 1 = (98−x²)/x²
2 (x²−18)² − x^4 = x² (98−x²)
2 (x²−18)² − x^4 = 98x² − x^4
2 (x²−18)² = 98x²
(x²−18)² = 49x²
(x²−18)² − 49x² = 0
(x²−7x−18) (x²+7x−18) = 0
(x+2) (x−9) (x−2) (x+9) = 0
x = ±2, x = ±9
Since x is diameter of circle, it must be longer than chords → x > 7
*x = 9*

[michaeldoerr]

The answer is 9 units. And it looks like this a good example of knowing what would be the right circle theorem to make use of in order to justify auxiliary lines. I am kind of wondering, how many circle theorems justify auxiliary lines???

[solimana-soli]

A fast solution can be drawing segment PQ on the right side simmetrically to AP and BQ in the middle getting an isosceles trapezoid with minor base 7 and lateral sides 3, major base x. We can find x with Euclid’s theorem on APB right triangle:
3^2=x*(x-7)/2
x^2-7x-18=0
X=9
Very nice problem, very nice the solution by Mathboost❤

[oscarcastaneda]

Knowing cos(theta) = 7/(2r) and then using the Law of Cosines with one of the triangles with two sides "r" and third side "3" one can determine that 2r is 9.

Theta is the angle subtended by the "3" side and the center of the semicircle.

[ludmilaivanova]

@ 6:53 connect A and Q and get a right triangle AQM (angle ACB=90 degrees) which is similar to BPM. from that, 6/a= 7+a/3. 18=7a+a^2. a=2. sinse triangle ABC is isosceles D= BM = 9.

[ekoi]

similar triangles: (a+7)/(6) = (3)/(a)
a=2
x=a+7 = 2+7=9

[kateknowles]

Adding -{lines AQ and}- BP to the diagram and labelling O the centre of AB ( X is the diameter of the semicircle and O is the centre.),
-{ then APB and AQB are two rightangled triangles with 90º angles at P and at Q respectively. AQ^2= X^2 - 7^2 and BP^2 = X^2 - 3^2 by Pythagoras' theorem.
AC.CQ is equal to BC.CP if we label the point of intersection of AQ and BP as C. A, P, Q and B are given points on a (semi-)circle.
(AC+CQ)^2 = AQ^2 = AC^2 +CQ^2 +2AC.CQ
(BC+CP)^2 = BP^2 = BC^2 + PC^2 + 2BC.CP Now BP^2 - AQ^2 is equal to BC^2 +PC^2 - AC^2 - CQ^2 +(2BC.CP- 2AC.CQ, this term in the bracket equals zero)}-
but I think there should be a solution that does not use AQ and does not use C

Let ½X = r the radius, which equals OA, OP, OQ and OB Now angles AOP and POQ are equal and naming them each tº then the value of angle QOB is 180º- 2tº
and 7x7 = 49 = r^2 +r^2 - 2.r.r cos(180-2t) and 3x3 =9 = r^2 +r^2 - 2.r.r cos (t)
49-9 = 2.r.r ( cos (t) - cos (180 -2t) r.r = r^2= (40/2) / ( cos(t) + cos (2t)) cos (2t) = cos^2(t)-sin^2(t) = 2cos^2(t ) -1
r^2 = 20 / (2cos^2(t) +cos(t) -1) the denominator factorises as ( 2c+1)(c-1) where c= cos(t)

So a couple of paths i have shown here which are to be abandonned and the solution to be sought in the video provided here.

Peeping at it now I perceive that I forgot that construction can be extended beyond the given shape, and this can simplify the solution.

Thank you again, Math Booster, well done to students who " got it" on their own efforts, and encouragement and better "luck" next time to others!

[bpark]

You can simplify this problem by rearranging the line segments with the 7 long one at the center. Then symmetry gives 2 triangles: one hypotenuse 3, one leg r-3.5, & the other hypotenuse r, one leg 3.5. Equating Pythagoras for the 2 other legs gives equation
3² - (r - 3.5)² = r² - (3.5)². 2r² - 7r - 9 = 0. r = 4.5. X = 2r = 9.

[timeonly]

Draw segments AQ & PB, and call their lengths c & d, respectively. ∠APB & ∠AQB are right angles (By Thales Thm).
So, applying Pythagorean Thm to ∆AQB & ∆APB: c² + 49 = x² and 9 + d² = x², so that c=√(x²-49) and d=√(x²-9).
All the vertices of quadrilateral APQB are on the semicircle (&, hence, the circle), so APQB is a CYCLIC quadrilateral.
So, the product of the diagonals is equal to the sum of the products of opposite sides (Ptolemy's Thm). Applying, we get:
(AQ)(PB) = (AP)(QB) + (PQ)(AB)
c d = (3)(7) + (3)x = 3(x+7)
Subbing in c and d from earlier:
√(x²-49) √(x²-9) = 3(x+7)
Squaring both sides and solving for x:
(x²-49)(x²-9) = 9(x+7)²
(x-7)(x+7)(x²-9) = 9(x+7)²
[x is a diameter which is the longest chord in the circle, so x > 7. Hence x-7 > 0. And because x-7 is not zero, we can divide both sides by it.]
(x-7)(x²-9) = 9(x+7)
x³ - 7x² - 9x + 63 = 9x + 63
x³ - 7x² - 18x = 0
x(x² - 7x -18) = 0
[x > 7 ⇒ x ≠ 0]
x² - 7x -18 = 0
(x-9)(x+2) = 0
x = 9 or x = -2 (elim., since x>7)
so x=9.
Done!!

[marioalb]

My solution, same as video.
Excellent !!!

[imetroangola]

*_Solução:_*

No triângulo retângulo ∆APB:

*sen θ = 3/x*

No triângulo retângulo AQB:

*cos 2θ = 7/x*

Como cos 2θ = 1 - 2sen² θ, então:

7/x = 1 - 2 × (3/x)²

Seja 1/x = y > 0. Daí,

7y = 1 - 18y² → 18y² + 7y - 1 = 0.

∆ = (-7)² - 4× (18)×(-1) = 121

y = (-7 + 11)/36 = 4/36 = 1/9

x = 1/y → *x = 9 unidades*

[tezchan_pmksr]

円の中心（ABの中点）を点Oとすると

よってPA:AQ=OB:BP　→PA×BP＝AQ×OB

3×√(x²-3²)=√(x²-7²)× x/2
9x²-81=(x⁴-49x²)/4
x⁴-85x²+324=0　→x²=4、81　→x=2、9
x＞7より、x=9

[marcgriselhubert]

We use an orthonormal center O, the center of the circle, and first axis (OA).
We have A(R; 0) B(-R; 0) P(R.cos(t); R.sin(t)) Q(R.cos(2.t); R.sin(2.t)) with R the radius of the circle and t = angleOAP, 0°<t<90°
Then VectorAB(R.(1 -cos(t); R.sin(t)) and knowing that 1 -cos(t) = 2.(sin(t/2))^2 and sin(t) = 2.sin(t/2).cos(t/2) we have AB = 2.R.sin(t/2)
VectorDC(R.(1+cos(2.t); R.sin(2.t)) and knowing that 1 +cos(2.t) = 2.(cos(t))^2 and sin( 2.t) = 2.sin(t).cos(t) we have DC = 2.R.cos(t)
So we have 2.R.sin(t/2) = 3 and 2.R.cos(t) = 7. By division we have cos(t)/sin(t/2) = 7/3, and knowing that cos(t) = 1 -2.(sin(t/2)^2 we get:
(1 -2.X^2)/x = 7/3 with X = sin(t/2), or 6.(X^2) +7.X -3 = 0. Delta = 121 = 11^2, so X = (-7 +11)/12 = 1/3 (the other solution is negative, so rejected). Finally X = sin(t/2) = 1/3. We had 3 = AB = 2.R.sin(t/2), so 3 = 2.R.(1/3) and R = 9/2 and AB = 2.R = 9.

[pwmiles]

Taking perpendiculars from the centres of the chords and forming right triangles

2 arcsin(3/x) + arcsin(7/x) = pi/2
2 arcsin(3/x) = pi/2 - arcsin(7/x)
Take sine of both sides
2 (3/x) sqrt(1-9/x^2) = sqrt(1 - 49/x^2)
36/x^2 (1-9/x^2) = 1 - 49/x^2
Multiply by x^4, rearrange and simplify
x^4 - 85x^2 + 36.9 = 0
x^2 = (85 +/- sqrt(85^2 - 36^2)) / 2
x^2 = (85 +/- 77) / 2
x^2 = 81 or 4
x>=7 so
x=9

[raghvendrasingh]

❤
By Ptolemy's theorem
3x+21 = sqrt {(x^2 - 49)(x^2 - 9) }
x^3 - 67x - 126 = 0
X= 9 by RRT

[shaozheang]

The answer is 14. Use the properties of cyclic quadrilaterals and the thales theorem to find out by connecting A to Q.

[skwest]

Or,
1) Connect AQ.
2) Drop perpendicular from P to AQ, intersecting AQ at R.
3) Extend PR through O to intersect circle at S, making PS a diameter (length X), with PR = a, and RS = X - a.
4) For Right△APR, calculate (by Pythagoras) the length of AR as √(3² - a²) = √(9 - a²).
5) Use Intersecting Chords (AQ and PS) to figure 'a' in terms of X... i.e. AR * RQ = PR * RS:
√(9 - a²) * √(9 - a²) = a * (X - a), or
9 - a² = aX - a², so that
9/X = a
6) Next, use Right △AQB and Pythagoras to establish that:
7² + [2 * √(9 - a²)]² = X², which is,
49 + 36 - 4a² = X²; simplified to 'a' in terms of X becomes:
√[(85 - X²)/4] = a
7) Substituting for 'a' from #5, we get:
√[(85 - X²)/4] = (9/X), square both sides to get
(85 - X²)/4 = 81/X², or
85X² - X⁴ = 324, or
X⁴ - 85X² + 324 = 0
8) Substituting Z = X² we have:
Z² - 85Z + 324.
9) Use the quadratic equation to solve for Z:
Z = [85 ± √(85² - 4 * 1 * 324)]/2, or
Z = [85 ± √(7225 - 1296)]/2
Z = [85 ± √(5929)]/2 = [85 ± 77]/2, or
Z = 81 or 4
10) Reverting with Z = X² we get:
X = √81 = 9, or X = √4 = 2

Obviously, X cannot be equal to 2, as the resulting △AQB would be degenerate.
Therefore, X = 9. Q.E.D.

[gzimatipi]

AQ= √3²+3²=√18=3√2
X=√ (3√2)²+7²
X=√9x2+49
X=√67
X=√9*7
X=3√7

[SGuerra]

🎉🎉🎉 A questão é muito boa. Eu resolvi aplicando o Teorema dos Cossenos e a minha solução não exige uma construção auxiliar. Parabéns pela escolha 🎉🎉🎉 BRASIL Novembro de 2024.

[tocamelosuevos]

Hay algo que no me cuadra. El problema planteado tiene infinitas soluciones si se resuelve con un compás y una regla, dependiendo del ángulo que arbitrariamente formemos entre los segmentos PA y PQ y, por ende, de la longitud del segmento AQ. Cuando un problema geométrico tiene una 'única' solución trigonométrica, tiene tambíen una 'única' solución gráfica. Y no es el caso.