An isosceles triangle ABC (AB = BC) is inscribed with a circle whose radius length is 9 cm.

An isosceles triangle ABC (AB = BC) is inscribed with a circle whose radius length is 9 cm. The tangent l to the circle, parallel to the line AC, intersects the sides AB and BC at points P and T, respectively. It is known that BP: PA = 1: 3. Calculate the perimeter of the APTC quadrilateral.

1. Triangles ABN and PBK are similar, therefore:
BN: BK = BA: BP = 4x / x = 4;
(y + 2R): y = 4;
y + 2R = 4y;
3y = 2R;
y = 2R / 3 = 2 * 9/3 = 6;
BO = y + R = 6 + 9 = 15.
2. By the Pythagorean theorem to the triangle BOM:
BM ^ 2 = BO ^ 2 – R ^ 2;
BM ^ 2 = 15 ^ 2 – 9 ^ 2 = 225 – 81 = 144;
BM = 12.
3. From the similarity of triangles BPK, BOM and BAN:
AN: R = BN: BM;
AN = 9 * 24/12 = 18;
PK: AN = x / 4x = 1/4;
PK = 1/4 * AN = 1/4 * 18 = 4.5;
x: y = BO: BM;
x = 6 * 15: 12 = 7.5;
AP = 3x = 3 * 7.5 = 22.5.
4. Perimeter of the APTC quadrilateral:
P = 2 (PK + AN + AP) = 2 (4.5 + 18 + 22.5) = 2 * 45 = 90 (cm).
Answer: 90 cm.