Dr. Ashani Dasgupta
Forum Replies Created
-
Posts
-
Here is a link to some sequential hints on this problem (I replaced some of the coordinate bashing by elementary geometry).
I attached the coordinate geometry computations anyway. But I strongly suggest that you try this on your own first. It is not hard. Use the steps indicated above.
If you are absolutely stuck then look into the attached pdf file.
This is a very interesting problem. The easy solution is using brute force.
Short Answer: 1.61 (approximately); that is the Golden Ratio
Long Answer:
Suppose A= (0,a), C = (0,0), B = (b, 0).
Then AH /BH = \( \frac {a^2} {b^2} \) (why? either use geometry or brute force computation using coordinate geometry).
The fact that HP is perpendicular to AQ provides with the following equation: $$ 1 + \frac{b^2}{a^2} = \frac{a^2}{b^2} $$
(How? Find inradius, find the coordinates of P and Q, compute slopes of HP and AQ, set the product of the slopes equal to -1)
Assume \( \frac{a^2}{b^2} = t \) and solve for t.
I will upload the computations if you are unable to do it.
Clearly \( a_n = \frac{n^2}{n!} = \frac{ n}{(n-1)!} = \frac{n-1 +1}{(n-1)!} = \frac{1}{(n-2)!} + \frac{1}{(n-1)!} \)
Hence \( \sum a_n = \sum \frac{1}{(n-2)!} + \sum \frac{1}{(n-1)!} = 2e \)
Is the question complete?
For example, if we plug in a = 1, then we will get (possibly complex) solutions of b.
Computation:
if a = 1
\( (1-b)(1 - \sqrt{b} ) = 1 \)
Expanding we get \( 1 - \sqrt{b} - b + b \sqrt{b} = 1 \)
This implies \( b \sqrt b = b + \sqrt b \Rightarrow b = \sqrt b + 1 \)
Set \( \sqrt b =x \) and solve the qudratic.
This process will work for any value of a.
This problem reminds me of a problem from Pre RMO 2017. If you expand the given expression you will get \( a \sqrt a + b \sqrt b - a \sqrt b - b \sqrt a = 1 \). The Pre RMO problem set the values of \( a \sqrt a + b \sqrt b = 183 \) and \( a \sqrt b -+ b \sqrt a = 182 \)