Olympiad Maths Prep

Track / Stage 6 / 81 of 400 #1081 of 2000

Problem 1081

National olympiad, first round
Geometry Difficulty 6.1 Prove it

Given a circle kk and the segment ABAB. The lengths of the tangent segments drawn from AA and BB to kk are aa and bb, respectively. Is it true that a+b>ABa + b > AB if and only if the segment ABAB has no common points with kk?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

The task only makes sense if neither AA nor BB falls inside the circle. If neither AA nor BB is on the circumference of the circle, then the statement is true; otherwise, it is not. If, for instance, AA is on the circumference and BB is outside the circle, then a=0,b>ABa=0, b>AB, so a+b>ABa+b>AB, but the segment ABAB has a common point with kk, so in this case, the statement is not true (Figure 1).

If both AA and BB are outside the circle, we need to examine the following three cases:

(i) ABAB is outside kk,

(ii) ABAB touches kk at a point EE,

(iii) ABAB intersects kk.

The statement of the problem is that in case (i), a+b>ABa+b>AB, while in the other two cases, a+bABa+b \leq AB. In all three cases, it is true that the two tangent segments drawn from AA and BB to kk are of equal length, so it is always sufficient to examine the tangent segment that is more favorably positioned for us.

In case (i), we can assume that one of the two tangent lines separates ABAB and kk (i.e., the circle and the segment are on different sides of the tangent line), while the other does not. In our diagram, the tangent from AA is the separating one. Then the intersection point CC of the two tangent lines will be an internal point of the tangent segment from BB, but not of the tangent segment from AA. The points will be arranged as shown in Figure 2. Using the notation from the diagram:

a+b=AEA+BEB=AEA+CEB+CB=AEA+CEA+CB=AC+CB a+b=AE_A + BE_B = AE_A + CE_B + CB = AE_A + CE_A + CB = AC + CB

However, by the triangle inequality, AC+CB>ABAC + CB > AB, so the statement of the problem is true in this case.

In case (ii) (Figure 3), a+b=AE+EB=ABa+b = AE + EB = AB, so the statement is clearly true in this case.

In case (iii), we draw the tangents whose points of tangency are separated from the center OO of the circle kk by ABAB (Figure 4). Let the intersection points of the radii EAOE_AO and EBOE_BO with ABAB be MAM_A and MBM_B, respectively. Then the points A,MA,MB,BA, M_A, M_B, B are arranged in this order, and MAM_A and MBM_B may coincide if OO is on ABAB, so a+b<AMA+MBBABa+b < AM_A + M_BB \leq AB.

This proves that the statement of the problem is true precisely when both AA and BB are outside the circle.

Based on the work of Gábor Nyul (Debrecen, Fazekas M. Gymnasium, 1st year)

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.