ABC is a right angled triangle with ∠ABC=90∘ and side lengths AB=24 and BC=7. A semicircle is inscribed in ABC, such that the diameter is on AC and it is tangent to AB and BC. If the radius of the semicircle is an improper fraction of the form a/b, where a and b are positive, coprime integers, what is the value of a+b?

Hmm. Place point D such that ABCD is a rectangle.

Let the center of the semi-circle be at point O on AC.

Draw another semi-circle centered at O, to form a complete circle. Since it is tangent to AB, it is also tangent to DC.

So, the diameter of the circle is 7, and the radius is 7/2, so a+b=9.

its wrong steve

199

To find the radius of the semicircle inscribed in triangle ABC, we can make use of the properties of tangents drawn to a circle from an external point.

Let's denote the center of the semicircle as O and the point of tangency on AB as D. Since the tangent to the semicircle at D is perpendicular to AB, we have a right triangle ADO, where DO is the radius of the semicircle.

In triangle ABC, we are given that AB = 24 and BC = 7. Since ABC is a right-angled triangle, we can use the Pythagorean theorem to find the length of AC.

AC² = AB² + BC²
AC² = 24² + 7²
AC² = 625
AC = 25

Now, let's consider triangle ADO. We know that AD = AC - CD. Since AC = 25 and CD is the radius of the semicircle, we have:

AD = 25 - CD

To find CD, we observe that AD is equal to the sum of the hypotenuse and one of the legs of the right-angled triangle ADO. Using the Pythagorean theorem, we have:

AD² = AO² + OD²
AD² = (AO + OD)²
AD² = (AC - CD + CD)²
AD² = AC²

Thus, we have:

(25 - CD + CD)² = 25²
625 - 50CD + CD² = 625
CD² - 50CD = 0
CD(CD - 50) = 0

From this, we can see that there are two potential values for CD: CD = 0 or CD = 50. However, CD cannot be 0 since it is the radius of the semicircle, and a semicircle cannot have a radius of 0. Therefore, CD = 50.

Substituting back into the equation AD = 25 - CD, we get:

AD = 25 - 50
AD = -25

Since lengths cannot be negative, we discard this solution and consider the other side of the triangle.

Let's denote the point of tangency on BC as E. Since DE is also the radius of the semicircle, we can apply similar reasoning as above.

Using the Pythagorean theorem on triangle BDE, we have:

BD² = BE² + DE²
24² = BE² + 50²
576 = BE² + 2500
BE² = 2500 - 576
BE² = 1924

Taking the square root of both sides, we find:

BE = √1924
BE = 2√481

Therefore, the radius of the semicircle is half of BE (since it's a semicircle):

Radius = 1/2 * BE = 1/2 * 2√481 = √481

The improper fraction is already in the simplest form, so we can directly conclude that a = 1 and b = 481.

Thus, the value of a + b is:

1 + 481 = 482

Therefore, the value of a + b is 482.