We have 2 independent solutions, stating that the answer is 1/3. So 1/9 must be wrong. We should correct the question later. Sorry for the trouble. - 8foxes

Yes, indeed. cos(α)= 1/3. I made an error in calculation. Here's the complete solution: Let the semi-circle be x^2+y^2=b^2 with the right top corner as (a,b). The tangent is ax1+by1=b^2 with U as (x1,y1). So x1^2+y1^2=b^2 which gives y1=b(b^2-a^2)/(b^2+a^2) = -(b-2b/3) giving us a=b√2. Now let 180-α=2Φ. From the figure,tan(Φ)=b/(b√2/2)= √2 or (sec(Φ))^2=1+2=3 which in turn gives (cos(Φ))^2=1/3 or cos(2Φ)=2/3 -1=-1/3 and hence cos(α) = 1/3 Ajit

One easier way of doing this is as follows: Name the rectangle ABCD starting with the top left hand corner and going anti-clockwise. Let AB=2b and BC=a as before. Let DU meet BC in E and let midpt. of UD be F. Now BE = EU = UF =FD = a/2 and thus EC=a/2 but DE = 3a/2 and hence cos(DEC)= 1/3. Now if O be the centre of the semi-circle then triangles OBE, OEU, OUF & OFT are all congruent and thus α = /_TFD = /_BOU = /_DEC. Hence etc. Wud that be correct? Ajit

your reasoning is indeed correct but you seem to consider BE=EC=a/2 as obvious. I can't find a decisive argument to support that, except by assuming cos(DEC)=1/3 which obviously goes into circles... Any better idea?

Here is probably the simpliest solution: (Name the rectangle ABCD starting with the top left hand corner and going anti-clockwise - As Ajit defined)

1. |AD|=|UD|=2e (they are the common tangents from the same point). 2. Let DU intersect BC in point P. 3. |UP|=|BP|=e (they are the common tangents from the same point). 4. |PC| = 2e-e = e (ABCD is a rectangle). 5. Connect point U to the center of the circle, point O. 6. m(/_UPC)=a (Try to see the identical deltoids (kites) sharing the same side OU. 7. cos(a)=cos(/_UPC) = e/3e = 1/3.

Option E: cos(α)= 1/9

ReplyDeleteWe have 2 independent solutions, stating that the answer is 1/3. So 1/9 must be wrong. We should correct the question later. Sorry for the trouble.

ReplyDelete- 8foxes

Yes, indeed. cos(α)= 1/3. I made an error in calculation. Here's the complete solution: Let the semi-circle be x^2+y^2=b^2 with the right top corner as (a,b). The tangent is ax1+by1=b^2 with U as (x1,y1). So x1^2+y1^2=b^2 which gives y1=b(b^2-a^2)/(b^2+a^2) = -(b-2b/3) giving us a=b√2. Now let 180-α=2Φ. From the figure,tan(Φ)=b/(b√2/2)= √2 or (sec(Φ))^2=1+2=3 which in turn gives (cos(Φ))^2=1/3 or cos(2Φ)=2/3 -1=-1/3 and hence cos(α) = 1/3

ReplyDeleteAjit

One easier way of doing this is as follows: Name the rectangle ABCD starting with the top left hand corner and going anti-clockwise. Let AB=2b and BC=a as before. Let DU meet BC in E and let midpt. of UD be F. Now BE = EU = UF =FD = a/2 and thus EC=a/2 but DE = 3a/2 and hence cos(DEC)= 1/3. Now if O be the centre of the semi-circle then triangles OBE, OEU, OUF & OFT are all congruent and thus α = /_TFD = /_BOU = /_DEC. Hence etc. Wud that be correct?

ReplyDeleteAjit

your reasoning is indeed correct but you seem to consider BE=EC=a/2 as obvious. I can't find a decisive argument to support that, except by assuming cos(DEC)=1/3 which obviously goes into circles... Any better idea?

ReplyDeletebleaug

BE =EU (equal tangents) and DU =DA (equal tangents). Since EU=UH=HD we, therefore, can say that EC = a/2 and ED=3a/2.

ReplyDeleteWhat's the difficulty here?

Ajit

Here is probably the simpliest solution:

ReplyDelete(Name the rectangle ABCD starting with the top left hand corner and going anti-clockwise - As Ajit defined)

1. |AD|=|UD|=2e (they are the common tangents from the same point).

2. Let DU intersect BC in point P.

3. |UP|=|BP|=e (they are the common tangents from the same point).

4. |PC| = 2e-e = e (ABCD is a rectangle).

5. Connect point U to the center of the circle, point O.

6. m(/_UPC)=a (Try to see the identical deltoids (kites) sharing the same side OU.

7. cos(a)=cos(/_UPC) = e/3e = 1/3.

-binary

ok. i was missing the DU=DA part. thanks

ReplyDeletebleaug