Poland Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods 

1st method: Triangle ∆DOB: OD² + OB² = BD² 1² + 2² = BD² BD² = 1 + 4 = 5 BD = √5 Mirror arc AB, radius OB, and segment DB about OA to create arc AE, radius OE, and segment DE, with the now semicircle having a diameter EB. As E and B are points on the diameter and C is a right angle on the circumference, CE is a valid line segment, and ∆BCE is a right triangle. ---- (1) ∆DOE is a construction drawn to mirror ∆DOB that shares line segment OD, so ∆DOE and ∆DOB are congruent triangles. As ∠BCE = ∠DOE = 90° and ∠CEB and ∠OED are the same angle, ∆BCE and ∆DOE are similar. Triangle ∆BCE: BC/EB = OD/DE BC/4 = 1/√5 BC = 4/√5 CE/BC = OE/OD CE/(4/√5) = 2/1 CE = 8/√5 The shaded area is equal to the area of ∆BCE minus the area of ∆EDB. Pink Triangle ∆BCD: A = BH/2 - bh/2 A = (8/√5)(4/√5)/2 - 4(1)/2 A = 16/5 - 2 = 6/5 sq units 2nd method: Start from [---- (1)] above to create the semicircle. Now mirror the semicircle about EB to create a full circle, and mirror radius OA to form radius OF and thus diameter AF. By the intersecting chords theorem, the products of the opposite sides of intersecting chords are equivalent, so as CE and AF are chords that intersect at D, AD•DF = CD•DE. Remember that DE = BD = √5 (construction). AD•DF = CD•DE 1(3) = CD(√5) CD = 3/√5 Pink Triangle ∆BCD: CD² + BC² = BD² (3/√5)² + BC² = (√5)² 9/5 + BC² = 5 BC² = 5 - 9/5 = 16/5 BC = √(16/5) = 4/√5 A = (3/√5)(4/√5)/2 = (12/5)/2 = 6/5 sq units

Good methods. I do like the similar triangles method.

There is a very simple and quick method. Do the semi-circle construction as in your second method. The area of triangle PDO is 1 which is also the area of triangle BDO. The area of triangle PDB is 2. Triangle PBC is similar to triangle PDO, therefore BC/CP = DO/PO = 1/2. Therefore PC=2*BC. Let BC = x. Therefore 4*r^2 = 4*x^2 + x^2 = 5*x^2. x^2 = (4*r^2)/5 The area of triangle PCB = x^2 = 16/5. Shaded area = 16/5 -2 = 6/5 = 1.2.

1²+2²=DB²→ DB=√5 → Si "E" es el simétrico de "C" respecto al eje AO→ ED=DC → Potencia del punto "D" respecto a la circunferencia =AD(DO+r)=ED*DB→ ED=1*(1+2)/√5=3√5/5 =DC → CB²=(√5)²-(3√5/5)²=80/25→ CB=4√5/5 → Área DCB =DC*CB/2 =(3√5/5)*(4√5/5)/2 =6/5. Gracias y saludos.

We can use Ptolemy's theorem. Suppose CD = x and BC = y. Then we get 2x+y=2√5. In addition, x² + y² = 5. Then we solve x=3/√5 and y=4/√5. The area is equal to 6/5.

I think that I like the first method better than the second method. And I think is faster. I am starting to wonder if there are problems that can easily be solved by both the first and second methods described here.

Bakıdan salamlar.Əla həll etdiniz.Təşəkkürlər.

A wonderful geometry problem to relax the mind after a day at work. It can be shown that if R is the given radius then A = 0.3 * R^2.

Another method: DOB+BCD=180º then BCDO is inscrptible (in another smaller circle). OD=1, DB=sqr 5, OC=radius=2.By Ptolomeus Theorem 2CD+CB=2sqrt(5). Also by Pythagoras CD²+BC²=5. Solving: CD=3/sqr(5), BC=4/sqr(5). Shaded area=CD*BC/2=(3*4/5)/2=6/5.

Math Booster estimated PD=√5 . Let the inscribed circle on the quadrilateral ODCB. PD⋅PC=P0⋅PB => √5⋅PC=2⋅4 => PC=8/√5 Now DC=PC-PD=8/√5-√5 => *DC=3/√5* If you draw the heigh OE in the right triangle POD we have : (OPD)=1/2 PD⋅0E=I/2⋅√5⋅OE => (OPD)=I/2⋅√5⋅OE (1) (POD)=1/2 OP.OE=1/2⋅2⋅1=>(POD)= 1 (2) (1),(2) => 1/2 √5⋅0E=1⇒0E=2/√5 In triangle PCB , OE joins the midpoints of the straight lines PC and PB , so BC=2⋅.0E=2⋅2/√5=4/√5 At last area of triangle BDC =1/2⋅DC⋅BC 1/2=3/√5⋅4/√5=6/5 *I could estimate OE more easily , by Pythagoras theorem in triangle BDC , but I did this solution in order to demonstrate the theorem of the means of line segments.*

how do you conclude that P is collinear with DC ? looks that way on the drawing, but I see no proof.

So... I'm going with 6/5. 1. Complete the circle. 2. Extend segment BD through to intersect the circle at C'. 3. Extend segment CD through to intersect the circle at B'. 4. Extend radius BO through to intersect the other side of the completed circle at, say, X, and extend radius AO through to intersect the circle at A' (segment AA' is a diameter.) 5. Since ∠BCD is a right angle, we know that B' and unknown point X are the same point. 6. Connect new points B' and C' to create new triangle △B'C'D. 7. By vertical angles, ∠CDB and ∠C'DB', and the fact that angles ∠BCB' and ∠BC'B' are both right angles, we can establish that △BCD and△B'C'D are similar. 8. Then, since triangles △BOD and△B'OD are congruent (SAS), segment BD equals segment B'D forcing similar triangles △BCD and△B'C'D to be congruent. 9. And now to the part that counts. Since the length of segment OD = 1 (i.e. ½ of the radius of 2), segment BD can be calculated (by Pythagoras) to be √5. 10. We now have 2 chords, ADA' and BDC', intersecting at D, with: AD = 1 DA' = 3 BD = √5, and DC' = a By the Intersecting Chords theorem: AD • DA' = BD • DC', or 1 • 3 = √5 • a and, so, a = 3√5/5. 11. Since △BCD is congruent with△B'C'D, segment CD = C'D = 3√5/5. 12. Finally, we can establish the length of leg BC by Pythagoras: BC² + CD² = BD², or BC² + (3√5/5)² = (√5)² BC = 4√5/5. (it turns out that △BCD is a 3-4-5 right triangle) 13. Anyway, we now have a base and height for △BCD, of 3√5/5 and 4√5/5. Area is given as ½bh, or: ½(3√5/5)(4√5/5) = ½•12•5/25 = 6/5. Voilá. Now to watch and see if I got it right. Cheers! - s.west

Posto i cateti a=DC,b=CB..è l'angolo α=ODC..per la legge del coseno..4=1+a^2-2acosα..e 4=4+b^2-4bcos(180-α),da quest'ultima si ottiene cosα=-b/4,la sostiisco nella prima 3=a°2+ab/2..la abbino alla a^2+b^2=5 (vedi disegno)..quindi dal sistema a=3/√5..b=4/√5..(a=3 non è accettabile)..Apurple=(3/√5)(4/√5)/2=6/5

You're from which country?

6/5