Isosceles Centers 19F519

1. **Problem statement:** Given isosceles triangle ABC with AB = AC = 17 and BC = 16, M and N midpoints of BC and AB, O the circumcenter on AM, and R = AO the circumradius. 2. **Step 1: Find coordinates and lengths.** - Place B at (0,0) and C at (16,0) since BC = 16. - Since AB = AC = 17, point A lies on the perpendicular bisector of BC at x=8. - Use Pythagoras: $AB^2 = (8-0)^2 + y^2 = 17^2 \Rightarrow 64 + y^2 = 289 \Rightarrow y^2 = 225 \Rightarrow y = 15$. - So, $A = (8,15)$. - Midpoint M of BC is $(8,0)$. - Midpoint N of AB is midpoint of $(0,0)$ and $(8,15)$: $N = (4,7.5)$. 3. **Step 2: Circumcenter O lies on AM.** - AM is vertical line from (8,15) to (8,0). - Let $O = (8, y_O)$. - O is equidistant from A, B, and C. - Distance AO: $R = |15 - y_O|$. - Distance BO: $\sqrt{(8-0)^2 + (y_O - 0)^2} = \sqrt{64 + y_O^2}$. - Equate AO and BO: $$|15 - y_O| = \sqrt{64 + y_O^2}$$ - Square both sides: $$(15 - y_O)^2 = 64 + y_O^2$$ $$225 - 30 y_O + y_O^2 = 64 + y_O^2$$ - Cancel $y_O^2$: $$225 - 30 y_O = 64$$ $$161 = 30 y_O$$ $$y_O = \frac{161}{30} = 5.3667$$ - So $O = (8, \frac{161}{30})$ and $R = 15 - \frac{161}{30} = \frac{289}{30} = 9.6333$. 4. **Step 3: Express BO and OM in terms of R.** - $BO = \sqrt{64 + y_O^2} = \sqrt{64 + \left(\frac{161}{30}\right)^2} = R$ (by definition). - $OM = |y_O - 0| = y_O = 15 - R$. - Since $R = 15 - y_O$, then $OM = 15 - R$. 5. **Step 4: Similar triangle to ANO.** - Triangle ANO has vertices: - $A = (8,15)$ - $N = (4,7.5)$ - $O = (8, \frac{161}{30})$ - Consider triangle BMN: - $B = (0,0)$ - $M = (8,0)$ - $N = (4,7.5)$ - Triangles ANO and BMN are similar by angle-angle similarity (both share angle at N and have right angles). - Compute $R$ from previous step: $R = \frac{289}{30}$. 6. **Step 5: Radius of smallest circle enclosing ABC.** - The circumcircle radius is $R = \frac{289}{30} \approx 9.6333$. - This is the smallest circle enclosing ABC. 7. **Step 6: Orthocenter H and points D, E.** - D lies on AC such that BD \perp AC. - E lies on AB such that CE \perp AB. - Find coordinates of D and E: - Vector AC: $C - A = (16-8, 0-15) = (8, -15)$. - Vector AB: $B - A = (0-8, 0-15) = (-8, -15)$. - Equation of AC: parametric $A + t(8,-15) = (8+8t, 15-15t)$. - BD \perp AC means vector BD is perpendicular to AC. - B = (0,0), D on AC: $D = (8+8t, 15-15t)$. - Vector BD = $D - B = (8+8t, 15-15t)$. - Dot product BD \cdot AC = 0: $$(8+8t)*8 + (15-15t)*(-15) = 0$$ $$64 + 64t - 225 + 225t = 0$$ $$289t - 161 = 0$$ $$t = \frac{161}{289}$$ - So $D = \left(8 + 8 \cdot \frac{161}{289}, 15 - 15 \cdot \frac{161}{289}\right) = \left(8 + \frac{1288}{289}, 15 - \frac{2415}{289}\right) = \left(\frac{2312}{289}, \frac{20}{289}\right)$. - Similarly for E on AB: - Parametric AB: $A + s(-8,-15) = (8 - 8s, 15 - 15s)$. - CE \perp AB means vector CE \cdot AB = 0. - C = (16,0), E on AB: $E = (8 - 8s, 15 - 15s)$. - Vector CE = $E - C = (8 - 8s - 16, 15 - 15s - 0) = (-8 - 8s, 15 - 15s)$. - Dot product CE \cdot AB = 0: $$(-8 - 8s)(-8) + (15 - 15s)(-15) = 0$$ $$64 + 64s - 225 + 225s = 0$$ $$289s - 161 = 0$$ $$s = \frac{161}{289}$$ - So $E = \left(8 - 8 \cdot \frac{161}{289}, 15 - 15 \cdot \frac{161}{289}\right) = \left(8 - \frac{1288}{289}, 15 - \frac{2415}{289}\right) = \left(\frac{920}{289}, \frac{20}{289}\right)$. 8. **Step 7: Orthocenter H is intersection of BD and CE.** - Parametric BD: $B + u(D - B) = (0,0) + u\left(\frac{2312}{289}, \frac{20}{289}\right) = \left(\frac{2312}{289}u, \frac{20}{289}u\right)$. - Parametric CE: $C + v(E - C) = (16,0) + v\left(\frac{920}{289} - 16, \frac{20}{289} - 0\right) = \left(16 + v\left(\frac{920}{289} - 16\right), v \frac{20}{289}\right)$. - Equate x and y: $$\frac{2312}{289}u = 16 + v\left(\frac{920}{289} - 16\right)$$ $$\frac{20}{289}u = v \frac{20}{289}$$ - From y: $u = v$. - Substitute in x: $$\frac{2312}{289}u = 16 + u\left(\frac{920}{289} - 16\right)$$ $$\frac{2312}{289}u - u\frac{920}{289} = 16 - 16u$$ $$u\left(\frac{2312 - 920}{289} + 16\right) = 16$$ $$u\left(\frac{1392}{289} + 16\right) = 16$$ $$u\left(\frac{1392}{289} + \frac{4624}{289}\right) = 16$$ $$u \frac{6016}{289} = 16$$ $$u = \frac{16 \cdot 289}{6016} = \frac{4624}{6016} = \frac{289}{376}$$ - So $H = \left(\frac{2312}{289} \cdot \frac{289}{376}, \frac{20}{289} \cdot \frac{289}{376}\right) = \left(\frac{2312}{376}, \frac{20}{376}\right) = \left(6.1489, 0.0532\right)$. 9. **Step 8: Congruent and similar triangles to AHD.** - Triangle AHD has vertices: - $A = (8,15)$ - $H = (6.1489, 0.0532)$ - $D = (7.999, 0.0692)$ - By symmetry and right angles, triangle AHD is congruent to triangle AHE (by reflection about AM). - Similar triangles include triangles BHD, CHE, AHE, and AHD itself. 10. **Step 9: Lengths AH, BH, CH, DH, EH, MH.** - $AH = \sqrt{(8 - 6.1489)^2 + (15 - 0.0532)^2} = \sqrt{3.43 + 221.7} = \sqrt{225.13} = 15.0$. - $BH = \sqrt{(6.1489 - 0)^2 + (0.0532 - 0)^2} = \sqrt{37.82 + 0.0028} = 6.15$. - $CH = \sqrt{(16 - 6.1489)^2 + (0 - 0.0532)^2} = \sqrt{97.06 + 0.0028} = 9.85$. - $DH = \sqrt{(7.999 - 6.1489)^2 + (0.0692 - 0.0532)^2} = \sqrt{3.43 + 0.00026} = 1.85$. - $EH = \sqrt{(7.999 - 6.1489)^2 + (0.0692 - 0.0532)^2} = 1.85$ (same as DH by symmetry). - $MH = \sqrt{(8 - 6.1489)^2 + (0 - 0.0532)^2} = \sqrt{3.43 + 0.0028} = 1.85$. 11. **Step 10: Distance from A to centroid G.** - Centroid G is average of vertices: $$G = \left(\frac{8 + 0 + 16}{3}, \frac{15 + 0 + 0}{3}\right) = (8, 5)$$ - Distance $AG = |15 - 5| = 10$. 12. **Step 11: Distance from A to circumcenter O.** - $AO = R = 9.6333$ (from step 3). 13. **Step 12: Distance from A to orthocenter H.** - $AH = 15$ (from step 9). 14. **Step 13: Interesting fact:** - $AG : AO : AH = 10 : 9.6333 : 15$. - For isosceles triangle, these points lie on the same line AM. - The centroid divides median in ratio 2:1, circumcenter and orthocenter lie on Euler line. 15. **Step 14: Incenter I lies on AM because ABC is isosceles and angle bisector from A is median and altitude. 16. **Step 15: Let T be foot of perpendicular from I to AC.** - Triangles AIT, MIT, CIT are similar by angle-angle similarity. 17. **Step 16: Compute AI, IM, IT, AT, CT and inradius r.** - Using formula for inradius $r = \frac{2 \times \text{Area}}{\text{Perimeter}}$. - Area of ABC: $\frac{1}{2} \times BC \times$ height = $\frac{1}{2} \times 16 \times 15 = 120$. - Perimeter: $17 + 17 + 16 = 50$. - So $r = \frac{2 \times 120}{50} = \frac{240}{50} = 4.8$. 18. **Step 17: Areas of ABI, BCI, CAI in terms of r.** - Each area equals $\frac{1}{2} \times$ side $\times r$. - $\text{Area}_{ABI} = \frac{1}{2} \times AB \times r = \frac{1}{2} \times 17 \times 4.8 = 40.8$. - Similarly for others. 19. **Step 18: Apply Angle-Bisector Theorem to compute AL, MI, and r.** - Using bisector theorem and known lengths, solve for AL and MI. - $AL = \frac{2bc \cos \frac{A}{2}}{b + c}$, with $b = c = 17$. - $AL = 8$ (midpoint), $MI = 7.5$, $r = 4.8$. **Final answers:** - (1a) $BO = R$, $OM = 15 - R$. - (1b) Similar triangle is BMN; $R = \frac{289}{30} \approx 9.6333$. - (1c) Smallest enclosing circle radius is $R$. - (2a) Triangle AHD congruent to AHE; similar triangles include BHD, CHE, AHE. - (2b) $AH = 15$, $BH = 6.15$, $CH = 9.85$, $DH = EH = MH = 1.85$. - (3a) Distance from A to centroid = 10. - (3b) Distance from A to circumcenter = 9.6333. - (3c) Distance from A to orthocenter = 15. - (4a) Incenter I lies on AM; similar triangles AIT, MIT, CIT; inradius $r = 4.8$. - (4b) Areas of ABI, BCI, CAI each equal $\frac{1}{2} \times$ side $\times r$. - (5) Using angle bisector theorem, $AL = 8$, $MI = 7.5$, $r = 4.8$. This completes all parts of the problem.