Poland Math Olympiad | A Very Nice Geometry Problem

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[quigonkenny]

Triangle ∆ABC:
AB² + BC² = CA²
7² + 24² = CA²
CA² = 49 + 576 = 625
CA = √625 = 25

Let r be the radius of circle P. Let E, F, and G be the points of tangency between circle P and AB, BC, and CA respectively.

As GA and AE are both tangent to circle P and intersect at A, then GA = AE = 7-r. As FC and CG are both tangent go circle P and intersect at C, then CG = FC = 24-r.

CA = CG + GA
25 = (24-r) + (7-r)
25 = 31 - 2r
2r = 31 - 25 = 6
r = 6/2 = 3

As ABCD is a rectangle, then AB = CD and BC = DA. Thus ∆ABC and ∆CDA are congruent. By extension, circles P and Q are also congruent.

Let T be the point where PT is parallel to BC and QT is parallel to CD. PT = 24-2r and QT = 7-2r.

Triangle ∆PTQ:
PT² + QT² = PQ²
(24-2r)² + (7-2r)² = PQ²
(24-6)² + (7-6)² = PQ²
PQ² = 18² + 1² = 324 + 1
[ PQ = √325 = 5√13 units ]

[bpark]

A simpler way to get radius is to use radius = triangle area/half-perimeter. Half perimeter = (7 + 24 + 25)/2 = 28. Area - (1/2)(7)(24) = 84. r = 84/28 = 3.

[andrewj]

Oh, yes. Thanks for the explanation. I couldn’t get there because I’d forgotten the fact that tangents from a single point on a circle have the same length. I got the same answer though by using the engineer’s approach using trigonometry. 😊

[wuchinren]

r=(7+24-25)/2=3
the centers of the two circles are (3, 3) and (21, 4)
x=the distance between (3, 3) and (21, 4)=sqrt((21-3)^2+(4-3)^2)=sqrt(325)=5*sqrt(13)

[epsom]

AC=25, BM=BN=(7+24-25)/2=3, P(3, 3), Q(21, 4)
PQ^2=18^2+1^2=325=5^2*13

[jimlocke]

Finding the radius of the circle with center P using trigonometry: Construct AP. Drop a perpendicular from P to AB and label the intersection as point N. Note that AP bisects <BAC. Let <BAC be designated 2Θ, then <PAN = Θ. From ΔABC, tan(<BAC) = tan(2Θ) = BC/AB = 24/7. The tangent double angle formula is tan(2Θ) = 2tan(Θ)/(1 - tan²(Θ)). Replace tan(2Θ) with 24/7 and tan(Θ) with z, then 24/7 = 2z/(1 - z²), multiply both sides by (1 - z²): 24(1 - z²)/7 = 2z which expands and simplifies to 12z² +7z -12 = 0 with roots z = 3/4 and z = -4/3. z = tan(Θ) must be positive because Θ is acute, so tan(Θ) = 3/4. Let the radius of the circle be r. Then PN = r and AN = 7 - r. From ΔANP. tan(Θ) = PN/AN = r/(7 - r). So, r/(7 - r) = 3/4, 4r = 3(7 - r), 4r = 21 - 3r and r = 3. Skip ahead to 7:55, where Math Booster uses r = 3 to find length PQ.

[oscarcastaneda]

Since r = 3, just solve for the hypotenuse of the right triangle with legs of lengths (24 - 2r) and (7 - 2r).
This leads to 5sqrt(13).

[alexniklas]

From point P we draw a segment PM=2r, perpendicular to AC. Let us connect the points M and Q. From ∆PMQ using the Pythagorean theorem we find PQ
AC=√(7^2+24^2)=25
r=(7+24-25)/2=3
X^2= (2r)^2+[AC-2(AB-r)]^2=
=6^2+17^2=325
X=√325=5√13

[ناصريناصر-سب]

We have AC=25 according to the Pythagorean theorem, from which r=(AB+BC-AC)/2=3, and we have PQ²=(7-2r)²+(24-2r)²=325, so PQ=5√13

[kateknowles]

Thank you for this..
I am sure that this took me rather a long time, but I plodded straight through (this way) with no errors, no corrections so am enjoying the moment of submitting the comment.

Drawing quadrants of circles UQV and SPT so that BTPS and DVQU are congruent squares with QU and SP parallel to AD and BC,
AS = 7-r = CU now extending SP and VQ to meet at a right angle at R,
QR = 7 - r - r and PR will have a value 24 -r -r (1) -> if r becomes known the hypotenuse PQ can be calculated using Pythagora' theorem.
Marking two more points along AC so that AW = AS and XC = AW = CU = 7 - r allows that angle XQU is supplementary with angle ACD tan(ACD) is 24/7 in a 7, 24, 25 triangle
WX = AC - (7 -r) - (7 -r) and ( 1/2) AC = will be 12 +1/2 by Pythagoras' theorem.
PWM will be a right-angle, if M is the midpoint of WX and of AC, and PM= 1/2 PQ
PW = r WM = 12 +1/2 -(7-r) PM.PM = PW.PW +WM.WM so there can be two equations for PQ in terms of r

PQ.PQ = QR.QR + PR.PR (1) (7-2r)(7-2r) = 49 -28r +4r.r (24-2r)(24-2r) = 576 -96r +4r.r PQ.PQ = 8r.r - 124 r + 625 (3)
PQ.PQ = 4 (PM.PM) = 4 (PW.PW + WM.WM) (2) 4PW.PW = 4r.r 4WM.WM = 4((25/2- 14/2)+r)(11/2+r) = (11+2r)(11+2r) PQ.PQ = 8 r.r + 44r + 121
solving these for r, and thence for PQ
44r +121 = 625 -124r as the 8r.r. terms cancel r =( 625-121 )/ (124+44) = 504 / 168 = (28.18)/(28.6) = 3
Using equation (3) PQ.PQ = 8. 3. 3 - 124. 3 + 625 = 625+72 - 372 =325 = 13 . 25
PQ is the square root PQ = 5 sqrt(13)

[cyruschang]

Diagonal length = √(7^2 + 24^2) = 25
7 - r + 24 - r = 25
r = 3
√(r^2 + (25/2 - (7 - r))^2) = √(9 + (17/2)^2) = (√325)/2
x = √325 = 5√13

[aulaseduardocurvello]

AB=25, but 25=(7-r)+(24-r), r=3, right triangle 8, 5×8, 5+3×3=(x×x)÷2, x=5square root13.

[marcgriselhubert]

Let's use an orthonormal center B and first axis(BC). We have A(0; 7) C(24; 0) VectorCA(-24; 7). Equation of (AC): (x).(7) - (y - 7).(-24) = 0
or 7.x + 24.y - 168 = 0. P is equidistant from (BC) and from (BA), so P(a; a) with a unknown real (0<a<7).
Distance from P to (AC) = abs(7.a + 24.a - 168)/sqrt(7^2 + 24^2) = abs(31.a - 168)/25. This distance is also equal to a, so we have:
31.a -168 = 25.a or 168 - 31.a = 25.a, and so a = 28 or a = 3. We reject a = 28, so a = 3 and then P(3; 3)
By symetry (the center of symetry is the middle of [A, C]) we have Q(24 - 3; 7 - 3), so Q(21; 4)
Finally VectorPQ(18; 1), PQ^2 = 18^2 + 1^2 = 325 and then PQ = sqrt(325) = 5.sqrt(13).

[zawatsky]

ΔАВС - пифагорова тройка 7:24:25, он равен ΔACD. Диаметр вписанной окружности R=7+24-25=7-1=6. Отнимаем его от катетов (они же - основание и высота прямоугольника), получаем проекции отрезка на них. Они же - декартовы координаты, они же - большой и малый катет прямоугольного треугольника, который можно построить от PQ. a=24-6=18, b=7-6=1.

[himo]

AC=√[7²+24²]=25 7-r+24-r=25 2r=6 r=3
PQ=x 7/2-3=1/2 24/2-3=9
9²+(1/2)²=(x/2)² 325/4=x²/4 PQ=x=√325=5√13

[RealQinnMalloryu]

(7)^2 (24)^2={49+576}=625/360=1.265 10^20^5^13 1^2^5^3 1^2^2^3^3 1^1^2^1^3 23 (ABCD ➖ 3ABCD+2).

[ivanhuertas]

you are confucing naming P and O to the same point !!! plus, p at the center of the circle in the rectangulo p at the tangent in the triangle p at the origen of the tangents

[ManojkantSamal]

Ans :little lesser than 12.02 (approximately )

[giuseppemalaguti]

Tre equazioni in tre incognite..r=3, x^2=325

[michagruszka]

Stop bullshitting the audience. It’s not Olympiad question. It’s a trivial case at the level of 6th-7th grade of primary school.