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.
+133 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.
+1632 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)
\(x + y = CM = 6\sqrt{2}\)
