Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods 

3 step solun just take height as x then apply Pythagoras to find side then apply cosine to find x. Then area = 1/2* 5 * x

I prefer the first method using trigonometric ratio and trigonometric identity with no construction needed. It can be more concisely presented as: 1. Let angle BAD be B and angle CAD be C such that B + C = 45 as given. Let height of triangle ABC i.e. AD be H. 2. Use definition of tangent ratio: tanB = 3/H and tanC = 2/H 3. Use compound angle identity for tangent ratio: tan(B+C) = (tanB + tanC)/(1 - tanB tanC) = tan 45 = 1 Hence [(3/H) + (2/H)]/[1 + (3/H)(2/H)] = 1 H^2 - 5H - 6 = 0 H = 6 or H = -1 (negative length rejected) 4. Area of triangle ABC = (1/2)(3+2)(6) = 15. The second method using construction of isosceles triangle which is similar to the original triangle is not that intuitive and it is more complicated to find the height AD from equations formulated by Pythagoras theorem and proportionality of corresponding sides of similar triangle.

Reflect ∆ADB about the side AB to form ∆AD'B. Similarly reflect ∆ADC about the side AC to form ΔAD''C. AD' =AD" = X and ∠D'AD"=90° and ∠AD'B =∠AD"C=90° Extend D'B and D"C to meet at E with ∠E=90°, consequently forming the Square AD'ED" with side X. The Right ∆BEC formed at the bottom, has sides (X-2, X-3, 5) which gives X=6 Hence area(∆ABC) =½*5*6=15

Hi! I have constructed a circumscribed circle of this triangle. Its radius is 2.5*sqrt(2). Further, it is not difficult to find the height of the triangle drawn from the vertex of this angle. It is equal to 6. Then the area of the triangle is 15. THANK YOU FOR THE BEAUTIFUL TASK!

It occured to me that the 45 degree angle at the apex might be divided by the same ratio as the two elements of the base stand relative to the total length of BC. That is to say 3 is 60% of the base, while 2 is 40% of the base. Applying that ratio to the 45 degree angle at the apex and you have 27 degrees for angle BAD and 18 degrees for angle CAD. Anyway, I get an area of 14.92. Which is pretty close to the 15 that the author has put out. Of course this could be sheer luck on my part, but I think there's a theorem somewhere that states this principle. Just can't remember it.

Right now I think that the second method is easier to understand and easier to intuit. Well tbf the first method was easy to understand and the second method is easier bc congruent and corresponding angles with a common corresponding and congruent angle result in a pair of triangles with the same letters as congruent corresponding angles being congruent. Easier to understand than the comments!

Nice second solution by Math booster . I solved it, using only Pythagoras theorem. But the system was a nightmare. I will post it tomorrow cause i am tired now 😊

I solved it by the 1st method....😊

*Outra solução:* Sejam AB=x, AC=y, AD= h e [ABC]= A. Usando a lei dos cossenos no triângulo ABC e sabendo que sin45°=cos45, temos: *25=x^2 + y^2 - 2xysin45° (1)* Os triângulos ABD e ADC são triângulos retângulos, logo pelo teorema de Pitágoras, temos: x^2=9 + h^2 e y^2=4 + h^2, somando membro a membro ambas expressões, obtemos: x^2+ y^2=13+ 2h^2, substituindo na expressão (1), temos: 25=13 +2h^2- 2xysin45° => *12=2h^2- 2xysin45° (2).* Observe que: A=xysin45°/2 => *xysin45°=2A* e h.5/2=A => *h=2A/5.* Substituindo tais expressões na expressão (2), temos o seguinte resultado: 12=16A^2/25 - 4A ÷(4) => 3=A^2/25 - A => 2A^2 - 25A-75=0. Daí, ∆=1225 => √∆=35, como A>0 então a única solução possível, segundo a fórmula de Bhaskara é: A=(25 + 35)/4 => *A=15.*

I extended BC and made your angle "alfa" 45 degrees. A big triangle area minus a newly constructed triangle area = the area under question. From the point B I dropped two perpendiculars-one on the CE, another on AC. The BCF triangle is a 5-4-3 straight triangle. Using the similarity of thre triangles I found all sides and hights. But my result was not 15 but 14. I did not check again the calculations, sorry. @ 9:54, sorry, I think the ratio should anclude not EC but EA. I wish anyone can check this, it is not obvious from the drawing.

So itonic that problems with geometry routinely have figures that are not to scale. Visual learners experience cognitive dissonance.

S(ABC)=15

Here is my solution Let BE⊥AC construction and AE=x ,EC=y. Notice that right triangle ABE is isosceles, so BE=AE=x Pythagoras theorem in right triangle ABE => AB²=x²+x² => AB=x√2 Also Pythagoras theorem in ΔBEC => x²+y²=25 (1) Express AD using Pythagoras theorem in triangles ADC, ABD : AD²=(x+y)²-4 and AD²=2x²-9 so (x+y)²-4=2x²-9 => …… => x²-xy=15 => y=(x²-15)/x. Now substitute y in (1) and you have : x²+((x²-15)/x)²=25 => ……. 2x⁴-55x²+225=0 Let ω=x² and we have 2ω²-55ω+225=0 => D=1125>0 ω=45/2 or ω=5 (is rejected) ω=45/2 => x²=45/2 We found AD²=2x²-9=> AD²=(2•45)/2-9 =>AD²=36 AD=6 and area of triangle (ABC)=(AD•BC)/2=(6•5)/2=15 s.u *ω=5 is rejected cause ω=5 => x²=5* *but y=(x²-15)/x

Why is this method incorrect: Area = ((5AD)/2) (half of base x height) Area = 1/2.AB.AC.sin45 [(in a triangle of sides a, b and c, Area = 1/2.b.c.sin(A), being A the angle opposed to the side a), right?] ((5AD)/2) = 1/2.AB.AC.sin45 5AD = AB.AC.(sqrt2/2) (10AD/sqrt2) = AB.AC 5.sqrt2.AD = AB.AC Get both sides to the power of 2 25.2.AD^2 = AB^2.AC^2 50AD^2 = AB^2.AC^2 By Pythagoras Theorem: AB^2 = 3^2 + AD^2 = 9 + AD^2 AC^2 = 2^2 + AD^2 = 4 + AD^2 Substituting in the equation: 50AD^2 = (9 + AD^2)(4 + AD^2) Let AD^2 be x 50x = (9 + x)(4 + x) 50x = x^2 + 13x + 36 37x = x^2 + 36 The only positive result for x is 1. AD^2 = x = 1 ===> AD = sqrt1 = 1 which is incorrect :( I actually don't know what I did wrong, could someone help me plsss?