A Very Nice Geometry Problem | 2 Different Methods

[@ناصريناصر-س4ب]

In triangle ABD cos(ADB)=(28²+26²-30²)/(2*28*26)=5/13 and according to Kashi's theorem in triangle ADC we have AC²=26²+35²-2*26*35*cos(ADC)=26²+35²-2*26*35*(-5/13)=2601 and from it AC=x=51

[@imetroangola17]

*Solução:* Seja a área [ABD]=L. Usando a fórmula de Heron no ∆ABD: p= (30 + 26+ 28)/2 = 42 L= [42.(42-30)(42-28)(42-26)]½ L = (42 × 12 × 14 × 16)½ L= (2×3×7×2²×3×2×7×2⁴)½ L= (2⁸×7²×3²)½ = 2⁴×7×3= *L = 336.* Por outro lado, seja h a altura do triângulo ABC traçada do vértice A até o ponto Q pertencente ao segmento BD. logo: h × BD/2 = L h= 336 × 2/ 28 → *h = 24.* Por Pitágoras no ∆AQD: QD² = AD² - h² = 26² - 24² = 10² QD=10. Por Pitágoras no ∆APC: x² = h² + PC² = h² + (QD + DC)² x² = 24² + (10 + 35)² x² = 576 + 2025 = 2601 x = 2601½ → *x = 51 unidades.*

[@raghvendrasingh1289]

cos ADB = (169+196-225)/(2×13×14) = 5/13 AM = 26×(12/13) = 24 DM = 10 MC = 45 , x^2 = 576+2025 = 2601 x = 51 second method x^2 = 26^2+35^2+2×26×35×(5/13) = 676+ 1225+700 = 2601 x = 51

[@davidellis1929]

There's an incredibly easy solution if you know your Pythagorean triangles, their multiples, and how they fit together. ABD is double the well-known 13-14-15 triangle, which can be viewed as a 9-12-15 right triangle (triple the Pythagorean 3-4-5) back to back with a Pythagorean 5-12-13 triangle, sharing a leg of length 12. Then the altitude AM from A to BC is double the length 12, or 24. Triangle AMD is double the 5-12-13, so MD=10. Now in right triangle AMC, AM=24 and MC=10+35=45, which are triple the legs of the well-known Pythagorean 8-15-17 triangle, so x=AC=3*17=51. This insight enabled me to solve the problem without writing anything!

[@AmirgabYT2185]

30²=26²+28²-2•26•28•cosα 1460-1456cosα=900 1456cosα=560 cosα=5/13 x²=26²+35²-2•26•35•cos(180°-α) cos(180°-α)=-cosα=-5/13 x²=1901-1820(-cosα) x²=1901-1820•(-5/13) x²=1901+700 x²=2601 x=51 Answer: 51.

[@jimlocke9320]

It can be solved by applying Heron's theorem three times. First, we apply Heron's theorem to ΔABD. Let sides a, b, c be 26, 28, 30. The semi-perimeter is s = (a + b + c)/2 = (26 + 28 + 30)/2 = 84/2 = 42. So, area is √(s(s-a)(s-b)(s-c)) = √(42(42 - 26)(42 - 28)(42 - 30)) = √(112896) = 336. Using CD as the base, ΔACD has the same height as ΔABD, but its base is 35/28 = 5/4 times as long, so it has area (5/4)(336) = 420. Apply Heron's formula with a, b, c = 35, 26, x. s = (x + 61)/2, s - a = (x - 9)/2, s - b = (x + 9)/2, c = 61 - x)/2. We find (s - a)(s - b) = (x² - 81)/4 and s(s - c) = (61² - x²)/4 = (3271 - x²)/4. So, √(s(s-a)(s-b)(s-c)) = √(((3271 - x²)/4)((x² - 81)/4)). Let m = x². Then, area = √(((3271 - m)/4)((m - 81)/4)) = √((-m² +3802m - 301401)/16) = (1/4)√(-m² +3802m - 301401). So, (1/4)√(-m² +3802m - 301401) = 420, √(-m² +3802m - 301401) = 1680, -m² +3802m - 301401 = 2822400, m² - 3802m + 3123801 = 0, (m - 2601)(m - 1201) = 0. So, x = √(2601) = 51 or x = √(1201) or approx. 34.66. Heron's formula is applied again to ΔABC to determine if either root is valid. ΔABC's area is ((28 + 35)/(28)) = (9/4)(336) = 756. x = 51 produces the correct area, x = √(1201) does not. So, x = 51, as Math Booster also found.

[@johnbrennan3372]

Area of triangle ABD=336( using formula). So half of |AM| by |BD| = 336 that is |AM|=24. Using pythag on triangle AMD, |MD|=10.Then using pyth. on triangle AMC, x=51.

[@WahranRai]

8:13 Stewart's theorem

[@michaeldoerr5810]

Looks like I shall look up Stewart's theorem and Kashi's theorem for the second method. Also the x = 51.

[@kateknowles8055]

Angle ADC = 180-angle ADB = 180- y (say). cosine y = (676 + 784- 900 )/(2 x 26 x 28) = 560/(56x26) = 5/13 -5/13 =( AD.AD+ DC.DC- X.X )/(2 x 26 x35) X.X = 676+1225 + 26x35x2x5/13 = 1901 + 700= 51x51 X is 51

[@kateknowles8055]

( Rewriting : minus 5/13 = (AD.AD +DC.DC minus X.X)/ (2x26x35) )

[@rabotaakk-nw9nm]

ΔABD=(26/28/30)=(13/14/15)2 cosB=(15²+14²-13²)/(2•15•14)=3/5 ΔABC=(x/63/30)=(x'/21/10)3 x'=vʼ(21²+10²-2•21•10•3/5)=17 AC=x=3x'=3•17=51 😁

[@appybane8481]

11:25 : x^2=9*289 x=3*17 x=51

[@wasimahmad-t6c]

35+10=45×45+24×24=2601sqwarooth 51

[@SGuerra]

A questão é muito boa. Parabéns pela escolha. Brasil Novembro de 2024.

[@giuseppemalaguti435]

ADB=α ..cos(α/2)=√(42*12/28*26)=√(9/13)...x^2=26^2+35^2-2*26*35cos(180-α)=676+1225+1820cos(2arccos(√9/13))=676+1225+1820(2*9/13-1)=676+1225+700=2601..x=√2601...boh, spero di non sbagliare...ah ah

[@ritwikgupta3655]

What is the name of the rule in the 2nd method? Very useful. I have not seen it before.

[@WahranRai]

Stewart's theorem

[@ritwikgupta3655]

@WahranRai thanks

[@davidellis1929]

You can prove Stewart's theorem by using the Law of Cosines on the two supplementary angles at the base of the interior segment.

[@nenetstree914]

51

[@yakupbuyankara5903]

X=51

[@nantesloire]

Meister es ist nicht x².4 sondern 4x² .... immer die Zahlen vor Parameter.