Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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- Опубліковано 16 чер 2024
- Math Olympiad | A Very Nice Geometry Problem | 2 Methods
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I prefer the first method using trigonometric ratio and trigonometric identity with no construction needed. It can be more concisely presented as:
Reflect ∆ADB about the side AB to form ∆AD'B. Similarly reflect ∆ADC about the side AC to form ΔAD''C.
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
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.
Outra solução:
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!
I solved it by the 1st method....😊
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.
Nice second solution by Math booster .
Here is my solution
So itonic that problems with geometry routinely have figures that are not to scale.
S(ABC)=15
Why is this method incorrect: