A Difficult Geometry Problem | Romania Math Olympiad Question

Рет қаралды 10,124

Күн бұрын

Пікірлер: 36

@marioalb9726 Жыл бұрын

Side of square 's'= 12 cm Diagonal 'd' of square : d² = 2 . s² = 2 . 12² d = 16,97 cm Circle inscribed in isósceles right triangle: 2 (s - r) = d 2s - 2r = d r = (2 s - d) /2 r = 3,515 cm Semicircle inscribed in isósceles right triangle: s + R = d R = d - s R = 4,97 cm Segment RS: = 'a' Subtracting in diagonal direction: s - a = s - r a = r Pythagorean theorem: X² = (R+r)² + r² X² = 8,484² + 3,515² X = 9,184 cm ( Solved √ )

@andydaniels6363 11 ай бұрын

Exactly my approach to the problem. It’s only difficult if one insists on solving it from scratch instead of building on known results.

@spacer999 Жыл бұрын

Since RMS and NSC are isolesces right angle triangles with 45deg side angles, RS must equal to a (radius of small circle) and SC must equal to b (radius of large circle). RS+SC = a+b = AC/2 = √2*s/2, where s=12. To find a, note that AP=AR=AC/2 --> s-a=√2*s/2 --> a=s(1-√2/2) Now (MN)²=(RS)²+(a+b)²=a²+(a+b)²=s²(1-√2/2)² + s²/2=s²(2-√2) MN = s*√(2-√2)

@اممدنحمظ Жыл бұрын

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

@jimlocke9320 Жыл бұрын

At 11:20, note that ΔCNS is a 45°-45°-90° right triangle, therefore isosceles (

@marioalb9726 Жыл бұрын

Side of square 's'= 12 cm Diagonal 'd' of square : d² = 2 . s² = 2 . 12² d = 16,97 cm Circle inscribed in isósceles right triangle: 2 (s - r) = d 2s - 2r = d r = (2 s - d) /2 r = 3,515 cm Semicircle inscribed in isósceles right triangle: s + R = d R = d - s R = 4,97 cm Pythagorean theorem: X² = [s-(R+r)]² + (s - r)² X² = (12-8.485)²-(12-3,515)² X = 9,184 cm ( Solved √ )

@Dimitar_Stoyanov_359 Жыл бұрын

[11:10] △CSN is a right isosceles triangle, so ⇒ CS = NS = b = 12 - 2a ⇒ b = 12 - 2·(12 - 6√2) ⇒ b = 12·(√2 - 1)

@Dimitar_Stoyanov_359 Жыл бұрын

[3:47] R - middle point for AC, therefore ⇒ AR = RC = 12 - a = (12√2)/2 ⇒ 12 - a = 6√2 ⇒ a = 12 - 6√2

@quigonkenny 4 ай бұрын

Let O be the point of tangency between circle M and AC and let P be the point of tangency between semicircle N and AC. Let r be the radius of circle M and R be the radius of semicircle N. Let E and F be the points of tangency between circle M and AB and BC respectively. Draw ME and MF. As radii of M, ME = MF = r. Additionally, ∠BFM = ∠MEB = 90°, as AB and BC are tangent to M at E and F. Thus ∠FME = 90° and MEBF is a square with side length r. EB and BF are tangents of circle M that intersect at B, so they are equal in length. If EB = BF, then AE must equal FC, as AB = BC = CD = DA = 12. FC and CO, and AE and OA, are two pairs of tangents to circle M that intersect at C and A respectively, and FC = AE, so CO and OA are equal as well. This means that O is at the midpoint of AC, and thus CO = OA = FC = AE = AC/2. AC is the diagonal of a side length 12 square, so AC = 12√2. This means that FC = 12√2/2 = 6√2. Since BC = 12 and BF = r, r = 12-6√2. DA and AP are two tangents of semicircle N that intersect at A, so they are equal. DA = AP = 12 and AC = 12√2, so PC = 12√2-12. NP is perpendicular to PC, as AC is tangent to semicircle N at P, so ∆NPC is a right triangle. As ∠PCN = 45° as AC is the diagonal of the square, then ∠CNP must equal 45° as well, and so ∆NPC is an isosceles right triangle and since NP = PC and NP = R, R = 12√2-12. Drop a perpendicular from M to T on CD. As MT is parallel to BC and ME = r, MT = 12-r = 12-(12-6√2) = 6√2. CT = r and ND = R, so NT = 12-r-R. NT = 12 - r - R NT = 12 - (12-6√2) - (12√2-12) NT = 12 - 12 + 6√2 - 12√2 + 12 NT = 12 - 6√2 Triangle ∆NTM: NT² + MT² = MN² (12-6√2)² + (6√2)² = MN² MN² = 144 - 144√2 + 72 + 72 MN² = 288 - 144√2 = 144(2-√2) MN = √(144(2-√2)) = 12√(2-√2) ≈ 9.184

@kinno1837 Жыл бұрын

Wow, it's just amazing😮 I hope you can make more videos like this.

@EddieDraaisma Жыл бұрын

A geometrical solution: M must lie on bisector of BAC, N on bisector of CAD. Angle NAM then is 45 deg. M also must lie on diagonal BD, so angle NDM is 45 deg. Then ADNM is a cyclic quadrilateral, thus angle ANM = angle ADM = 45 deg. So triangle ANM is right-angled isosceles triangle, thus NM = AM = AN / sqrt(2). AN = AD / cos(22.5 deg), cos(22.5 deg) = sqrt((2 + sqrt(2))/4), NM = 12 * sqrt(2 - sqrt(2))

@zsboya Жыл бұрын

Hello! I understand up to the first 2 sentences, and it is clear that ADM is 45 degrees. But what makes ANM also 45 degrees?

@PS-mh8ts 11 ай бұрын

@@zsboyaFirst he proves that angle NAM is 45°. Then he observes that M lies on the diagonal, and hence angle NDM is also 45°. Consider a circle that passes through A, M, and N. Thus MN is a chord of the circle. A, which lies on the circumference of the circle subtends an angle 45°. But D also subtends an angle 45° with NM. Thus, we prove that D must also lie on the same circle that passes through A, M, and N. In short: AMND is a cyclic quadrilateral. Now, consider the chord AM. It subtends two angles, one at D, and another at N, and they too ought to be the same. But ADM is 45° (because he proves that NDM is 45°). Thus, ANM must also be 45°. Thus, in triangle ANM, both angles A and N are 45° each, making it a right-triangle, right-angled at M. The solution follows thence.

@PS-mh8ts 11 ай бұрын

Excellent solution! 💯👍

@CanalMiTube Жыл бұрын

I did similar. All diagonals are solved directly by multiplying by √2. Greetings.

@n.662 Жыл бұрын

This is beautiful solution, how found radius! 👍

@markkinnard796 Жыл бұрын

Thumbnail didn't specify that ABCD is a square. Two incorrect thumbnails in a row.

@advancedintention7169 Жыл бұрын

Thanks sir....

@luigipirandello5919 Жыл бұрын

Very Nice problem. Thank you.

@niki2669 8 ай бұрын

dont get me wrong, but this is 7th grade math problem Olympiad in Romania

@alexutz6450 Жыл бұрын

Hello from Romania 🎉

@octobre4623 10 ай бұрын

How can we decide if this exercise is difficult or not ? (I am 75 and found the correct answer quite quickley).

@ТатьянаЕфименко-ц5р Жыл бұрын

Спасибо.

@giuseppemalaguti435 Жыл бұрын

R=raggio cerchio maggiore DN... r=raggio cerchio minore.. Risulta MN=sqrt[(12-r)^2+(12-R-r)^2] devo perciò calcolare R e r... R+sqrt2R=12(in basso a destra).. R=12/1+sqrt2...in alto a sinistra risulta tg22, 5=r/12-r=sqrt2-1...r=12-6sqrt2