Germany Math Olympiad | A Very Nice Geometry Problem

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[bpark]

Easier way to find AP: Draw line through point P parallel to MN, intersects QN at point R. Triangle PRQ is right triangle with hypotenuse PQ = 3, short leg QR = 1, so long leg RP = √8 = 2√2. Triangle AMP is congruent to this triangle PRQ so AP =3.
Height of triangle ABC = 3 + 1 + 2 + 2 = 8. Triangle ASC is similar to triangle PRQ so by scaling SC = 2√2 & base BC = 4√2. Area = (1/2)(4√2)(8) = 16√2.

[ekoi]

equations of the two circles
x^2+(y-2)^2=2^2
x^2+(y-5)^2=1^2

perpendicular line formula
d = (Ax+By+C)/sqrt(A^2+B^2)
let tangent line be:
y=mx+b
-mx+y-b=0
mx-y+b=0

let d=r, A=m, B=-1, and C=b
r = (mx-y+b)/sqrt(m^2+1)


substitute (0, 2) and r=2 to r = (mx-y+b)/sqrt(m^2+1).
2 = (m(0)-2+b)/sqrt(m^2+1)

substitute (0, 5) and r=1 to r = (mx-y+b)/sqrt(m^2+1).
1 = (m(0)-5+b)/sqrt(m^2+1)

system of 2 equations
b=8
m=sqrt(8) or - sqrt(8)

choose m=- sqrt(8).
y=-sqrt(8)x+8

x intercept:
y=0
0=-sqrt(8)x+8
x=sqrt(8)

x is half the base

y intercept:

x=0
y=-sqrt(8)(0)+8
y=8

y is the height

Area of the triangle:

1/2 A = 1/2 * (sqrt(8)) * (8)= 16sqrt(2)

[Rudepropre]

16*sqrt(2). Used similarity: x/1 = (x+2)/3 so x = 3 and thus height of ABC = 8. Used similarity again 8/y = sqrt(32)/2. 2y*8 = area of ABC: so [ABC] = 16*sqrt(2).

[Giovanni-ewsj]

Another way to calculate height: S ∞ = a1/1-q = diameter of the first circle/(1-diameter of the second/diameter 1) ∴ = 4/(1/2) = 8

[RealQinnMalloryu]

(1)^2.(2)^2={1+4}=5 2^15.2^15 1^3^5 1^3^5 1^2^3.1^2^3 2^1.1^3 2.3(ABC ➖ 3ABC+2).

[quigonkenny]

Let O be the center of the larger circle and P the center of the smaller. Let M and N be the points of tangency between circle O and AB and CA respectively. Let S be the point of tangency between circle O and BC and T the point of tangency between the two circles.

Draw DE, where DE is parallel to BC and goes through point T. By symmetry, as both circles are fully inscribed, T is the midpoint of DE and S is the midpoint of BC. As the point of tangency between two circles and the centers of those circles are always collinear, then AS passes through P, T, and O. Let V be the point between A and T where AS passes through the circumference of circle P, opposite T.

As DE and BC are parallel, then ∠ADT = ∠ABS, as corresponding angles, and similarly ∠TEA = ∠SCA as those are also corresponding angles. As S is the midpoint of BC, and ∠OSC = 90°, as OS is a radius of circle O and BC is tangent to circle O at S, then AS is a perpendicular bisector of triangle ∆ABC, and thus ∆ABC is an isosceles triangle, and ∆ADE is a similar isosceles triangle.

As ∆ABC and ∆ADE are similar triangles, and circles O and P respectively are identically inscribed in those triangles, then the triangles and their respective circles share the same ratio of similarity to each other. AT and AS, as perpendicular bisectors of the respective triangles, also share that ratio.

OS/AS = PT/AT
2/AS = 1/AT
AS = 2AT

As AS = 2AT, then T is the midpoint of AS and thus D and E are the midpoints of AB and CA respectively. AD = DB, CE = EA, and AT = TS = OT+OS = 2+2 = 4. AS = AT+TS = 4+4 = 8.

Draw OM and ON. As AM and NA are both tangent to circle O and intersect at A, then AM = NA, and as OM and ON are radii of circle O, ∠AMO = ∠ONA = 90°. As OA is common, ∆AMO and ∆ONA are congruent right triangles. OA = AS-OS = 8-2 = 6.

Triangle ∆AMO:
AM² + OM² = OA²
AM² + 2² = 6²
AM² = 36 - 4 = 32
AM = √32 = 4√2

As S is the perpendicular bisector of the base BC of isosceles triangle ∆ABC, then ∆BSA and ∆ASC are congruent right triangles. As ∠AMO = ∠BSA = 90° and ∠SAB = ∠OAM, then ∆AMO and ∆BSA are similar, as are ∆ONA and ∆ASC.

OM/AM = BS/AS
2/4√2 = BS/8
(4√2)BS = 2(8) = 16
BS = 16/(4√2) = 2√2

BC = 2BS = 2(2√2) = 4√2

Triangle ∆ABC:
A = bh/2 = (4√2)(8)/2 = 4(4√2)
[ A = 16√2 ≈ 22.627 sq units ]

[zawatsky]

Высоту можно найти и более изощрённым, алгебраическим, способом - это сумма бесконечной убывающей арифметической прогрессии. )

[santiagoarosam]

Trazamos el segmento PR de igual longitud y paralelo a MN y obtenemos el triángulo rectángulo QRP de lados: QR=2-1=1; QP=2+1=3 y RP=√(3²-1²)=2√2---> El triángulo PMA es congruente con QRP y el CSA es semejante---> SA=2+3+3=8---> SA/SC=RP/QRH---> SC=2√2---> Área ABC =8*2√2 =16√2 u².
Gracias y saludos

[zdrastvutye]

10 print"mathbooster-german math olympiad":dim xm(1), ym(1), x(2), y(2), r(1)
20
30 130
40
50 dg=dgu1+dgu2:return
60 xs=xm(1)-r(1):gosub 40
70 40:if dg1*dg>0 then 70
80 xs=(xs1+xs2)/2:gosub 40:if dg1*dg>0 then xs1=xs else xs2=xs
90 if abs(dg)>1E-10 then 80 else return
100 gosub 60:
110
120
130 gosub 100
140 100:if df1*df>0 then 140
150 xa=(xa1+xa2)/2:gosub 100:if df1*df>0 then xa1=xa else xa2=xa
160 if abs(df)>1E-10 then 150:rem interpolieren um die hoehe zu berechnen ***
170 ar:rem Stop
180 lh=xm(0)-xa:goto 200
190 xbu=x*mass:ybu=y*mass:return
200
210
220 masx<masy then mass=masx else mass=masy
230 xba=0:yba=0:for a=1 to 3:ia=a:if ia=3 then ia=0
240 x=x(ia):y=y(ia):gosub 190:xbn=xbu:ybn=ybu:goto 260
250 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
260 gosub 250:next a:for a=0 to 1:x=xm(a):y=ym(a):gosub 190:circle xbu, ybu, r(a)*mass
270 next a:adr=lh*y(2):print "die flaeche=";adr
mathbooster-german math olympiad
23.7796754
die flaeche=23.7796754
>
run in bbc basic sdl and hit ctrl tab to copy from the results window

[luizcamargo]

Please, deactivate automatic translation in this channel, it's awful!