Let AB be a diameter of a circle, and let C be a point on the circle such that AC=8 and BC=4 The angle bisector of ACB intersects the circle at point M Find CM.

Guest Dec 28, 2022

#1**+1 **

Here we go:

Because they have the same associated arc,

Let the intersection of AB and CM be P.

Since

That means AP/CP = PM/PB = AM/CB. Since we know CB, we have AP/CP = PM/PB = AM/4.

Note that CM = CP+PM.

By Pythagorean theorem, AB = sqrt(80).

By the angle bisector theorem, AP/8 = BP/4 and AP+BP = sqrt(80). Solving we get AP = 8 sqrt(5)/3 and BP = 4 sqrt(5)/3. If only SSA similarity was a thing...

Now we input these new values to get: 8 sqrt(5)/(3CP) = PM/(4 sqrt(5)/3) = AM/4. However, we can get CP as well with a little more calculation.

Let's drop a perpendicular from C to PB. Call the intersection point K. We have PK + BK = 4 sqrt(5)/3, PK^2 + CK^2 = CP^2, BK^2+CK^2=16, and finally (8sqrt(5)/3+PK)^2+CK^2=64.

Solving, we get CP = 8sqrt(2)/3. Thus we input and get PM = 10sqrt(2)/3. Adding, we get **6sqrt(2)=CM.**

As you can see, my solution involved lots of formulas (inscribed angle theorem, similar triangles, Pythagorean theorem, angle bisector theorem, etc) and tedious bashing... there's probably a better way. Follow along at your own risk.

tinfoilhat Dec 28, 2022

#2**+3 **

Answer is at bottom for the "i dont care about explanation" peoples.

heres a few law of cosines mixed with basic algebra, remember the principle of moving one of the crazy radicals to the other side! :)

Let the center of the circle be O. We have two triangles ACO and BCO with those angles being 45 degrees each. Let CO = x. We also know that ABC is a right triangle with hypotenus AB because of inscribed angle theorem, so AB = square root(8^2 + 4^2) = 4sqrt(5). Let OM = y.

By the law of cosines, we have OA^2 = 8^2 + x^2 - 16x(cos45) => OA^2 = 64 + x^2 - 8x[sqrt(2)]. We also have OB^2 = 4^2 + x^2 - 8x(cos45) = 16 + x^2 - 4x[sqrt(2)]. Since OA + OB = 4sqrt(5), and OA = \(\sqrt{64+x^2 - 8x\sqrt{2}}\), and OB = \(\sqrt{16 + x^2 - 4x\sqrt{2}}\). Then we substitute in OA and OB, subtract OB from both sides, square both sides: We need to solve for x, or OC, so we can use power of a point w/ similar triangles to get OM and add OC and OM.

\(\sqrt{64+x^2-8x\sqrt{2}}=4\sqrt{5} - \sqrt{16+x^2-4x\sqrt{2}}\)

\(64 + x^2 - 8x\sqrt{2} = 80 + (16 + x^2 - 4x\sqrt{2}) - 8\sqrt{5(16 + x^2 - 4x\sqrt{2})}\)

\(64 + x^2 - 8x\sqrt{2} = 96 + x^2 - 4x\sqrt{2} - 8\sqrt{80 + 5x^2 - 20x\sqrt{2}}\), combine like terms then divide both sides by 4.

\(8\sqrt{80 +5x^2 - 20x\sqrt{2}} = 4x\sqrt{2} + 32\)

\(2\sqrt{80 + 5x^2 - 20x\sqrt{2}} = x\sqrt{2} + 8\), square both sides.

\(4(80 + 5x^2 - 20x\sqrt{2}) = 2x^2 + 64 + 16x\sqrt{2}\), distributive property, combine like terms.

\(320 + 20x^2 - 80x\sqrt{2} = 2x^2 + 64 + 16x\sqrt{2}\)

\(18x^2 - 96x\sqrt{2} + 256 = 0\), divide both sides by 2, then use quadratic formula.

\(9x^2 - 48x\sqrt{2} + 128 = 0\)

\(x = {48\sqrt{2}\pm\sqrt{(48\sqrt{2})^2-(4)(9)(128)}\over18} = {48\sqrt{2}\pm0\over18} = {8\sqrt{2}\over3}\) DANG THAT LOOKS NICE!

Plugging x into OA's and OB's equations, we have:

\(OA = \sqrt{64 + {128\over9} - {128\over3}} = \sqrt{320\over9} = {8\sqrt{5}\over3}\) and

\(OB = \sqrt{16 + {128\over9} - {64\over3}} = \sqrt{80\over9}={4\sqrt{5}\over3}\), ratio is accurate by angle bisector theorem. These are neat numbers. NOW FOR:

BY POWER OF A POINT (remember that x = OC), OA * OB = OC * OM.

\(({8\sqrt{5}\over3})({4\sqrt{5}\over3})=({8\sqrt{2}\over3})y\)

\({160\over9} = {8y\sqrt{2}\over3}\)

\({20\over3\sqrt{2}}=y\)

\(y = {10\sqrt{2}\over3}\)

And since CM = x + y, here is our answer: 6sqrt(2)

\(x + y = CM = 6\sqrt{2}\)

proyaop Dec 28, 2022