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Difficulty: HardPythagorean Theorem and Special Right Triangles

A chord ABAB of length 2424 is drawn in a circle with center OO and radius 1313. A point PP lies on the circle such that the area of triangle ABPABP is maximized. What is the perimeter of triangle ABPABP?

  1. A
    24+81324 + 8\sqrt{13}
  2. B
    12+121312 + 12\sqrt{13}
  3. 24+121324 + 12\sqrt{13}Answer
  4. D
    24+61324 + 6\sqrt{13}
  5. E
    24+121724 + 12\sqrt{17}

Answer

The perimeter of triangle ABPABP is 24+121324 + 12\sqrt{13}.
The distance from center O to chord AB forms a 5-12-13 right triangle with half the chord length (12) and the radius (13), giving OM = 5. To maximize triangle area, P must be on the major arc, making the altitude PM = 13 + 5 = 18. Using the Pythagorean theorem in right triangle AMP with legs 12 and 18 gives AP = sqrt(12^2 + 18^2) = 6sqrt(13). The total perimeter is base AB plus twice AP, which equals 24 + 12sqrt(13).

Step-by-Step Solution

1
Find the distance from the center OO to the chord ABAB.
Let MM be the midpoint of chord ABAB. Since AB=24AB = 24, AM=12AM = 12. Triangle OMAOMA is a right triangle with hypotenuse OA=13OA = 13 and leg AM=12AM = 12. By the Pythagorean theorem: OM=132−122=169−144=25=5OM = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5.
The perpendicular from the center of a circle to a chord bisects the chord and forms a right triangle with the radius.
2
Determine the location of PP that maximizes the area of △ABP\triangle ABP and calculate the altitude.
The area of △ABP\triangle ABP is 12×AB×h\frac{1}{2} \times AB \times h, where hh is the perpendicular distance from PP to segment ABAB. Area is maximized when hh is maximized. Point PP must lie on the major arc along the diameter perpendicular to ABAB. Thus, the maximum height is PM=PO+OM=13+5=18PM = PO + OM = 13 + 5 = 18.
Since the base ABAB is fixed, maximizing the area requires maximizing the height perpendicular to ABAB.
3
Calculate the length of side APAP (and BPBP) using the Pythagorean theorem.
In right triangle AMPAMP, AM=12AM = 12 and PM=18PM = 18. Therefore: AP=122+182=144+324=468=613AP = \sqrt{12^2 + 18^2} = \sqrt{144 + 324} = \sqrt{468} = 6\sqrt{13}. By symmetry, BP=AP=613BP = AP = 6\sqrt{13}.
The perpendicular bisector of a chord creates two congruent right triangles for any point PP lying on it.
4
Calculate the total perimeter of △ABP\triangle ABP.
\text{Perimeter} = AB + AP + BP = 24 + 6\sqrt{13} + 6\sqrt{13} = 24 + 12\sqrt{13}.
The perimeter of a triangle is the sum of all three side lengths.

Key Concept

Pythagorean Theorem and Special Right Triangles in Circle Geometry
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Topics

Pythagorean Theorem and Special Right Triangles

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