Olympiad Mathematics | A Very Nice Geometry Problem | 2 Methods
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NIce way to begin the day : )
Using Heron's formula I found the area of △ABD to be 336 and its height to side BD to be 24. From this I focused on the right triangle with sides 24, 45, and AC from which AC = 51.
Already 16 👍s in 22 hours from Math Booster viewers for Heron's formula use here!
We put
Law of cosines twice
In triangle ABD to calculate angle B
In triangle ABC to calculate side length x
You can do it in one line by equating the cos of an angle to minus the cos of the angle of its supplement.
I call it the cos (supplement) rule.
Applying Stewart's Theorem in triangle ABC, we find 30²*35+x²*28=63*[26²+28*35], and from this x=51
Ese es el 2do método que ha realizado en el video.
@@Eror7403Yes, when I found that the solution in the video was using Stewart’s method, I suggested another method, which is available in the comments. Thank you.
@@Eror7403 We do not re-demonstrate known theorems already demonstrated (otherwise demonstrate Pythagoras at each use)
Let's assume that
The problem can be solved easily by recognizing multiples of Pythagorean right triangles with integer sides and integer areas, specifically 3-4-5, 5-12-13 and 8-15-17. The 26-28-30 triangle is double the well known 13-14-15 triangle, whose altitude to the side of length 14 divides it into 9-12-15 (three times 3-4-5) and 5-12-13. So the altitude to the side of length 28 is 24 units and divides it into segments of length 18 and 10. The altitude forms another right triangle whose legs are 24 and 45 (10+35), three times 8 and 15, so x is three times 17, or 51.
Awesome, I thought along the same line of thought : )
You can use Herron's formula to get area [ABD], and since 1/2 * (altitude) * (base), you can get the altitude straight away: Altitude = ( [ABD] * 2 ) / [BD]. Then find [a] via Pythagoras on [altitude] and [AD], then find [x] via Pythagoras on [altitude] and [a] + [DC].
We think Alike : )
Stewart's theorem:
b²m+c²n=(m+n).(d²+mn)
x².28+30².35 = (28+35).(26²+28 .35)
x = 51 cm ( Solved √ )
Cosine rule:
x²=30²+(28+35)²-2.30.(28+35)cosα
cosα= (30²+63²-x²)/3780
Cosine rule:
26²=30²+28²-2.30.28.cosα
cosα= (30²+28²-26²)/1680
Equalling:
(30²+63²-x²)/3780=(30²+28²-26²)/1680
4869-x² = 9/4 *1008
x²= 2601 --> x = 51cm (Solved √)
Stewarts theorem. Other than that pythagoras or even cosine rule
51
Draw a perpendicular line inside the left triangle (ABC). Let the height = AP
Let BP = n, then CP = 28-n
Hence, AP^2 =30^2 - n^2
But AP^2 , also= 26^2 - (28-n)^2
Hence, 30^2 - n^2 = 26^2 - (28-n)^2 since both =AP^2
30^2 -26^2 + 28^2 = n^2 - n^2 + 56n
2601 = 56 n
n=18
Hence, 28- n =10
Hence, the middle triangle is a 5-12-13 right triangle scaled up by 2
Hence, AP = 24 ( 2*12)
Hence, AP^2 +45^2 =x^2 note 45 comes from 35 + 10 . Recall that 28-n =10
Hence, 24^2 + 45^2 =x^2
2601 = x^2
51 = x
Hello Devon buddy. Thank you for writing this clear method. It shows that Math Booster leads the way for assorted rigorous solutions and successful followers.
@kateknowles8055 no problem glad it was helpful
cosine (ADB) = (BD.BD + AD.AD - AB.AB) /(2.BD.AD)
= (784 + 676 - 900) / (2.728) = (484 +76) / (16. 7.13) = 560/(16.7.13) = 5/13 = - cosine (ADC)
X.X = 26.26 + 35.35 - 2.26.35 cosine (ADC) =676 + 1225 - (52.35 )( - 5/13) = 1901 + 4.35.5 = 2601
The measure of X is 51 It needed the use of the cosine rule in each way, finding an angle, then finding a length.
Now I am going to see the video and comments.
I was thinking similarly but as this was happening I realized it was possible to do the area via Heron's formula without using pencil and paper so I opted for that route.
Good thinking my genius Math Amiga ==> MathMiga : )
In Method #1, let AM = h. From ΔADM, h² + a² = (26)² = 676. From ΔABM, h² + (28 - a)² = (30)². Expand: h² + (28)² - (2)(28)(a) + a² = (30)², h² + 784 - 56a + a² = 900. Simplify: h² + a² -56a = 116. Replace h² + a² by 676: 676 -56a = 116, -56a = -560, a =10. Replace a² by (10)² = 100 in h² + a² = 676: h² + 100 = 676, h² = 576 and h = 24. From ΔACM, x² = h² + (a + 35)° = 576 + (10 + 35)² = 576 + (45)² = 576 + 2025 = 2601, x = 51, as Math Booster also found.
*3° Método:*
No ∆ABD:
s=(30+26+28)/2 = 42. Pela fórmula de Heron:
[ABD]=[42×(42 -26)×(42 -28)×(42 -30)]½
[ABD]=[42×16×14×12]½=336
Seja H em BP, o pé da perpendicular do vértice A. Assim,
AH × BD/2 = 336 → AH × 28/2 = 336
AH = 336/14 = 24. Por Pitágoras, no ∆ABD:
HD² = AD² - AH² = 26² - 24² =100
HD = 10. Por Pitágoras, no ∆AHC:
AC² = AH² + HC²
x² = 24² + (10 +35)² =
x² = 2601
*x = 51.*
AからBDに垂線AH
HDの長さをyとすると、
26^2-y^2=30^2-(28-y)^2
28^2-56y=900-676=224
784-224=56y
560=56y
y=10
x^2=676-100+45^2
=576+2025
=2601
=51^2
∴x=51
ABC = δ → cos(δ) = 3/5 → x = 51
ADB = γ ACD = π - γ
Carnot theorem
cos γ = (28² + 26² - 30²) / 2∙26∙28 cos γ = 5 / 13
cos (π - γ) = - cos γ
x² = 35² + 26² + 2∙35∙25∙5/13 x = 51
Very nice problem with at least 5 solutions. The most suitable, in my opinion, with Stewart's Theorem.
it is a system of 2 nonlinear equations:
10 print "mathbooster-olympiad mathematics feb2025:dim x(1,2),y(1,2)":dim x(2,2),y(2,2)
20 la=30:lb=26:lc=28:ld=35:lh=(la^2-lb^2+lc^2)/2/lc:h=sqr(la^2-lh^2)
30 lx=sqr(h^2+(lc+ld-lh)^2):print lx:x(0,0)=0:y(0,0)=0:x(0,1)=lc:y(0,1)=0
40 x(0,2)=lh:y(0,2)=h:x(1,0)=lc:y(1,0)=0:x(1,1)=lc+ld:y(1,1)=0:x(1,2)=x(0,2):y(1,2)=y(0,2)
50 masx=1200/(lc+ld):masy=850/h:if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window.
Can you please provide proof of
b^2m+c^2n=a(d^2+mn)? Oh! I didn't remember Stewart's theorem.Ok.Now it is clear.Thanks.
Watch this video
ua-cam.com/video/ZfYG92_3LUQ/v-deo.html
See the article on _Stewart’s theorem_ in the English Wikipedia for a simple proof using the law of the cosines.
CAN BE DONE WITH COSINE RULE
Aから直線BCに垂線を引いて交点をEとし、DEの長さをaとする。
ピタゴラスの定理より30^2₋(28₋a)^2₌26^2₋a^2
900₋784₊56a₋a^2₌676₋a^2 116₊56a₌676 56a₌560 a₌10
x^2₌26^2₋10^2₊(35₊10)^2₌2601 X₌√2601₌51
X=51
Calculators are allowed?
ADB=α...cosα/2=√(42*12/26*28)=3/√13...(Briggs)...x^2=26^2+35^2-2*26*35cos(180-α)=676+1225+52*35(2*9/13-1)=676+1225+700=1925+676=2601
x=51
(28)^2(30)^2={784+900}=1684 {60°A60°B+60°C}=180°ABC 1684/180°ABC214 10^20^14 10^2^10^7^7 5^5^2^5^5^3^4^3^4 2^3^2^3^2^2^3^2^3^1^2^2^1^2^2 1^1^1^1^1^1^1^1^3^1^1^1^2 3^1^1^2 32(ABC ➖ 3ABC+2).(26)^2 (35)^2={576+1225}=1801 {45°A+45°B+90°C}=180°ABC 1801/180=10.1 5^5.1 2^3^2^3.1 2^1^1^3.1 2^1^3.1 2^3 (ABC ➖ 3ABC+2).
51
51