 # Week 15

Example 1. The sum of squares of the digits of a three-digit positive number is 146, while the sum of the two digits in the unit’s and the ten’s place is 4 times the digit in the hundred’s place. Further, when the number is written in the reverse order; it is increased by 297. Find the number.

Example 2. Let ABC be a triangle and D be a point on the segment BC such that DC = 2BD. Let E be the mid-point of AC. Let AD and BE intersect in P. Determine the ratios BP/PE and AP/PD.

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Solution 1. Let the number be abc i.e., 100a + 10b + c whose digits are a, b and c.

According to the symmetry of the question,

• a2 + b2 + c2 = 146;                   (1)
• b + c = 4a;                               (2)
• and 100c + 10b + a – 100a – 10bc = 297.                        (3)

From relation (3), we get

• 99(ca) = 297
• or, ca = 3
• or, c = a + 3                 (4)

Now using (4) in (2), we find that

• b + a + 3 = 4a
• or, b = 3a – 3              (5)

Finally, substituting the values from (4) and (5) in (1), we have

• a2 + (3a – 3)2 + (a + 3)2 = 146
• or, a2 + 9a2 – 18a + 9 + a2 + 6a + 9 = 146
• or, 11a2 – 12a – 128 = 0

Only acceptable value a from here is 4.

Hence, b = 3a – 3 = 3 × 4 – 3 = 9 and c = a + 3 = 4 + 3 = 7.

Therefore, the required number is 497.

Solution 2. Let F be the midpoint of DC, so that D, F are points of trisection of BC.

Now in triangle CAD, F is the mid-point of CD and E is that of CA.

Hence CF/FD = 1 = CE/EA.

Thus EF || AD. Hence we find that EF || PD.

Hence BP/PE = BD/DF. But BD = DF. We obtain BP/PE = 1.

In triangle ACD, since EF || AD we get EF/AD = CF/CD = 1/2.

Thus AD = 2EF. But PD/EF = BD/BF = 1/2.

Hence EF = 2PD. Therefore, AD = 2EF = 4PD.

This gives AP = ADPD = 3PD.

We obtain AP/PD = 3.