A Difficult Geometry Problem | Romania Math Olympiad Question

A Difficult Geometry Problem | Romania Math Olympiad Question

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[marioalb]

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 √ )

[اممدنحمظ]

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

[jimlocke]

At 11:20, note that ΔCNS is a 45°-45°-90° right triangle, therefore isosceles (<NSC = 90°, <SCN = 45° because diagonal and side of square meet at 45°, and by sum of Δ angles = 180°, this leaves <SNC = 45°), so b = 12 - 2a. Drop a perpendicular from CD. Call the intersection point W. Length NW = 12 - a - b = 12 - (12-2a) = a. Length MW = length QC = 12 -a. From Pythagorean theorem, (Length MN)² = (Length NW)² + (Length MW)² = a² + (12 - a)² = a² + 144 - 24a + a² = 2a² - 24a + 144. We note that a = 12 - 6√2 = 6(2 - √2), so a² = 36(4 - 4√2 + 2) = 72(3 - 2√2). Therefore: (Length MN)² = (2)(72)(3 - 2√2) - (24)(6)(2 - √2) + 144 = 432 - 288√2 - 288 +144√2 + 144 = 288 - 144√2 = 144(2 - √2). So, take the positive square root and length MN = 12√(2 - √2), as Math Booster also found.

[spacer]

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 + s²/2=s²(2-√2)
MN = s*√(2-√2)

[marioalb]

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 √ )

[kinno]

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

[Dimitar_Stoyanov_]

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

[n.]

This is beautiful solution, how found radius! 👍

[CanalMiTube]

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

[Dimitar_Stoyanov_]

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

[quigonkenny]

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

[EddieDraaisma]

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))

[markkinnard]

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

[octobre]

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

[giuseppemalaguti]

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,

[JEEIIT]

On 3rd September I am gonna give IOQM which is toughest in India, , ,

[niki]

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

[racquelsabesaje]

12+12+12+12= 48 m 12x4=48 m 2