Sweden Math Olympiad Problem | A Very Nice Geometry Challenge | 2 Different Methods To Solve

Sweden Math Olympiad Problem | A Very Nice Geometry Challenge | 2 Different Methods To Solve
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[@MathBooster]

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[@buziga236]

Hi sir! I'm sorry, but can you subscribe to calculus booster with your 2 other accounts? It just makes the travelling to your new channel way easier. Thanks, no problem if you don't

[@Fatihexioglu]

Congratulations for the New channel. This is great news.

[@Irtsak]

A variation in solving the exercise using *Stewart’s theorem* . Use *Math’s booster* shape in 5:11. Draw the straight segment AC. From orthogonal triangle ADC using Pythagoras theorem : AC²=45 In triangle BCE apply Stewart’s theorem : BC²⋅AE+EC²⋅AB=BE⋅(AC²+AB⋅AE) x²⋅5+10²⋅5=10⋅(45+5⋅5) 5x²+500=10⋅70 x²+100=140 x=2√10

[@d.m.7096]

Shorter Geometry solution - Draw DF || BC such that F is between E and A on side AE. Hence, BF = 6 and AF = 1. Let EF = ED = y Hence in right angle ∆ ADE, ED^2 + AD^2 = EA^2 Hence, y^2 + 3^2 = (y + 1)^2 Solving this gives y = 4 Draw AK || to BC such that K is between D and C. Then ∆ ADK is right angle ∆ and by Pythagoras theorem, AK = √10 ∆EAK ~ ∆EBC Hence, AK / BC = EA/ED = 5/10 = 1/2 Therefore, BC = X = 2√10

[@MarieAnne.]

At 6:50, instead of using law of cosine formula: cosθ = (a² + b² − c²) / (2ab), I would instead use the even better known law of cosine formula: c² = a² + b² − 2ab cosθ Applying this to △BEC, we can calculate x directly: BC² = BE² + CE² − 2(BE)(CE) cos α x² = 10² + 10² − 2(10)(10) (4/5) = 200 − 160 = 40 x = 2√10

[@scully-oh9we]

Extend top line to left by 3, then join this point to bottom left. This gives 3,4,5 triangle making new extended top angle 90°. Now extend new line of 4 by a further 2 downwards. This now forms 3 sides of a 6 sided square, so x is simply sqrt(2^2 + 6^2).

[@quigonkenny]

Extend BA and CD to intersect at E. As ∠ABC = ∠BCD = θ, ∆CEB is an isosceles triangle, and CE = EB. Let DE = y. CD + DE = EA + AB 6 + y = EA + 5 EA = y + 6 - 5 = y + 1 ∠CDA = 90° and CE is a straight line, so ∠ADE is 90° as well and ∆ADE is a right triangle. Triangle ∆ADE: 3² + y² = (y+1)² 9 + y² = y² + 2y + 1 2y = 8 y = 4 As AD = 3 and DE = 4, ∆ADE is a 3-4-5 Pythagorean triple right triangle and EA = 5 (confirmed as y+1 = 5). Thus CE = EB = 6+4 = 5+5 = 10. The law of cosines can be ised to determine x, as cos(∠E) = DE/EA = 4/5. BC² = CE² + EB² - 2CE(EB)(4/5) x² = 10² + 10² - 2(10²)(4/5) x² = 200 - 800/5 = 200 - 160 = 40 x = √40 = 2√10

[@jimlocke9320]

Variation 1st Method avoiding trigonometry: At 5:00, drop a perpendicular from B to CE and label the intersection as point F. Note that ΔADE and ΔBFE are similar. Therefore, by corresponding sides being in the same ratio, FE/DE = BF/AD = BE/AE = 10/5 = 2. BF is twice as long as AD, so has length 6. FE is twice as long as DE, so has length 8, leaving length CF = 2. Consider right ΔBCF. From the Pythagorean theorem, x² = 6² + 2² = 36 + 4 = 40, and x = √(40) = (√4)(√(10)) = 2√(10), as Math Booster also found.

[@sena23]

Yes this is the solution that i made in only 7-8 seconds maximum, even at the age of 42 :) he made a video of 19 minute length!

[@RAG981]

At 14 min, FC = 10cos theta, and you know cos theta = 1/rt10 from triangle ADE, so FC = 10/rt10 = rt10. So x = 2rt10. You went on for another 5 minutes doing some very strange methods.

[@spafon7799]

I think this is an easier way: Instead of making a parallelogram, run a line straight down from A to BC. Define J as the point they intersect. Run a line straight down from E to to BC. Define K as the point they intersect. Run a horizontal line from A to DC as in one of the solutions and name the intersection point E as in that solution. Now triangle ABJ and ECK are congruent, so EC equals 5. Therefore DE=6-5=1. By pythagorean applied to triangle ADE, segment AE equals sqrt(10). Now note that triangles ADE, AJB, and EKC are all similar. Thus BJ/AB=CK/EC=1/sqrt(10) Thus BJ=AB/sqrt(10)=5/sqrt(10)=sqrt(10)/2 and CK=EC/sqrt(10)=5/sqrt(10)=sqrt(10)/2. Then x= BJ+JK+KC= sqrt(10)/2+sqrt(10)+sqrt(10)/2= 2*sqrt(10)

[@robertlynch7520]

Actually, a completely different way works, too. Draw vertical from A → BC. Draw horizontal from A → DC Call it P Draw vertical from P → BC. The P→BC line is exactly same length as A→BC squares, congruency, and all that. Look at the upper '3' line AD. Since there are same-angles at B & C, then the intercept P is at side-length 5. With respect to the '3' triangle at top, it already has right angle, so hypotenuse is 𝒋² = 1² ⊕ 3² 𝒋² = 1 ⊕ 9 𝒋² = 10 𝒋 = √10 That's good, because it gives the length of A→P Next, need to find length of 𝒌, the distance from B to intersect with A→BC line (above). This turns out to be pretty easy too: H² + (𝒋 + 𝒌)² = √45² = 45 We know that H² = 5² - 𝒌² … expanded becomes 5² - 𝒌² + 𝒋² + 𝒌² ⊕ 2𝒋𝒌 = 45 … consolidate 𝒋² ⊕ 2𝒋𝒌 = 20 … but we know 𝒋 = √(10) so 10 + 2𝒌√10 = 20 2𝒌√10 = 10 𝒌 = 10 ÷ 2√10 𝒌 = 1.581139 And length of 𝒙, the base line is L = 𝒌 + 𝒋 + 𝒌 L = 1.581139 + √(10) + 1.581139 L = 6.3246… Which is yet another creative way to find the solution! ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

[@misterenter-iz7rz]

6-1=5, so sqrt(9+1)=sqrt(10), x^2+9=(x+1)^2, 2x=8, x=4, so 5, sqrt(10), 5, is upper isosceles triangle, the whole isosceles triangles should be 10, 2×sqrt(10), 10, thus x=2sqrt(10).

[@giuseppemalaguti435]

Le equazioni sono 45=4^2+x^2-8xcosθ(teorema di Carnot)..4/sin(θ-arctg1/2)=√45/sinθ(teorema dei seni)...trovo θ=arctg3/2,lo metto nella prima equazione e risulta x=29/√13..scusa ho scambiato 4 col 5...x=2√10

[@user-uc2qy1ff2z]

AB=5 BC=x CD=6 DA=3 ABC=BCD DCA=90 E is on CD. AE is parralel to BC AE²=(CD-AB)²+AD² EAB=AEC= 180-AED BC= AE+2 AB sin(EAB-90)= AE-2ABcos(EAB)=AE+2ABcos(AED)= sqrt(AE²)+2AB (CD-AB)/sqrt(AE²)= ((CD-AB)²+AD²+2AB (CD-AB))/sqrt(AE²)=((CD-AB)(CD+AB)+AD²)/sqrt(AE²)= (CD²-AB²+AD²)/sqrt((CD-AB)²+AD²)= (36-25+9)/sqrt(1+9)=20/sqrt(10) BC=2sqrt(10)

[@niteshmalhotra6162]

Hello Sir!! I regularly watch your videos on Geometry problems which I find quite informative!! Kudos to you!! In addition I would request you to kindly upload videos on algebra as well which captures a major chunk of the pie in competitive exams..

[@hadibiasl]

China 🇨🇳 Math Olympiad - Geometry - Find the length Y? kzbin.info/www/bejne/Z3-Uc6yYe86qbac

[@MathBooster]

Thank you. I will upload algebra problems on my second channel math hunter.

[@niteshmalhotra6162]

Thanks sir!! I really appreciate your efforts..

[@birandkoray]

for 2nd solution it is BF = FC, is it a coincidence?

[@MathBooster]

Yes

[@PremjitTalwar]

Too long and too convoluted. I solved it in less than a minute by drawing a horizontal line A parallel to BC, and intersecting DC at a point F. From F drop a line perpendicular to BC intersecting it at G. Geometry gives you DF =1. Pythagoras gives you AF. Similar triangles gives you the length GC. Now drop a line from A perpendicular to BC intersecting at H. BH = GC. X= 2 BH+HG (HG= AF).

[@MarieAnne.]

That's the method I used. But it's pretty similar to the second method in this video, except he uses a parallelogram, and an isosceles triangle, which is then divided into two right triangles similar to triangle ADF, instead of one rectangle and two right triangles similar to triangle ADF (which your method and mine did). I think the main reason you find his solution convoluted is that he is very wordy, and must describe each phase of the solution in a way that everyone can understand what he is doing.

[@dickroadnight]

AB = CE = 5, DE = DC - EC = 1 In Triangle, ADE, by pythag, AE = √10 = BF ********* θ = AED = acs(1/√10) (AE ||el BC) MC = EC cosθ = 5 cosθ = 5/√10 = √10/2 FC = 2 * MC = √10 x = BF + FC = 2√10