1. In triangle PQR let X be the intersection of the angle bisector of angle P with side QR, and let Y be the foot of the perpendicular from X to side PR. If PQ = 9, QR = 10 ,and PR = 17 then compute the length of XY.
2. The vertices of triangle ABC lie on the sides of equilateral triangle DEF, as shown. If CD = 5, CE = BD=2, and angle ACB = 90˚, then find AE.
3. In triangle ABC, point X is on side BC such that AX = 13, BX=10, CX = 4, and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC.
4. Let be the circle centered at with radius The lines and divide into four regions , , , and . Let denote the area of region . If
then find .
4. Let R be the circle centered at (0,0) with radius 10. The lines x=6 and y=4 divide R into four regions R1, R2, R3, and R4. Let [R_i] denote the area of region R_i. If [R1]>[R2]>[R3]>[R4]
then find [R1]-[R2]-[R3]+[R4].
-Guest
1.
Since X is the intersection of the angle bisector of angle P with side QR, then QX:XR = PQ:QR = 9:10. Also, since Y is the foot of the perpendicular from X to side PR, then XY:YP = QR:PR = 10:17. Combining these two ratios, we get QX:XY:YP = 9:10:17. Let QX = 9x, XY = 10x, and YP = 17x. Then by the Pythagorean Theorem on triangle PQX, we have PQ^2 = QX^2 + XQ^2 = 26x^2. Since PQ = 9, then 26x^2 = 9^2 = 81. Then x^2 = 9, so x = 3. Therefore, XY = 10x = 30.
2.
Since △DEF is equilateral, we have DE=DF=5.
Let x=AE. Then AF=5−x.
By the Law of Cosines on △DCB,
\begin{align*} BC^2 &= CD^2 + CB^2 - 2CD \cdot CB \cos 60^\circ \ &= 5^2 + 2^2 - 2 \cdot 5 \cdot 2 \cdot \frac{1}{2} \ &= 7. \end{align*}By the Law of Cosines on △CEA,
\begin{align*} AC^2 &= AE^2 + CE^2 - 2AE \cdot CE \cos 60^\circ \ &= x^2 + 2^2 - 2x \cdot 2 \cdot \frac{1}{2} \ &= x^2 - 2x + 4. \end{align*}By the Law of Cosines on △ABC,
\begin{align*} AB^2 &= AC^2 + BC^2 - 2AC \cdot BC \cos ACB \ &= (x^2 - 2x + 4) + 7 - 2(x^2 - 2x + 4) \cdot \cos 90^\circ \ &= x^2 - 2x + 11. \end{align*}Since AB=AF+BC=5−x+7, we have
[5 - x + \sqrt{7} = \sqrt{x^2 - 2x + 11}.]Squaring both sides, we get
[25 - 10x + 7 + x^2 - 2x + 4 = x^2 - 4x + 12.]Solving for x, we find x=5/3.
3.
Since the circumcircles of triangles ABX and ACX have the same radius, then triangle AXB is similar to triangle ACX. Let r be the radius of the circumcircles of triangles ABX and ACX. Then the circumradius of triangle ABC is also r. By Power of a Point,
[AX^2 + BX^2 = r^2] and [AC^2 + CX^2 = r^2]. Adding these equations, we get
[279 = 3r^2]. Then r=93
The area of triangle ABC is
[\frac{1}{2} \cdot BC \cdot r = \frac{1}{2} \cdot 10 \cdot \sqrt{93} = \boxed{5 \sqrt{93}}].
4.
The areas of regions R1, R2, R3, and R4 can be calculated as follows:
Region R1: This region is a sector of the circle with angle measure 90 degrees. The area of a sector with angle measure θ and radius r is 360∘θπr2, so the area of region R1 is 360∘90∘π(10)2=25π.
Region R2: This region is a triangle with vertices (0,0), (6,0), and (6,4). The area of this triangle is 21⋅6⋅4=12.
Region R3: This region is a triangle with vertices (0,0), (0,4), and (6,4). The area of this triangle is 21⋅0⋅4=0.
Region R4: This region is a rectangle with dimensions 6 by 8. The area of this rectangle is 6⋅8=48.
Therefore, the area of region R1 is 25π, the area of region R2 is 12, the area of region R3 is 0, and the area of region R4 is 48. Hence,
[R1] - [R2] - [R3] + [R4] = 25π - 12 - 0 + 48 = 25π + 36.
In conclusion, if [R1]>[R2]>[R3]>[R4] then [R1]-[R2]-[R3]+[R4] = 25π + 36.
+6 Consider ΔABC. Through each vertex draw a line perpendicular to the corresponding angle bisector. These three lines will form a triangle - say ΔA'B'C'. Note that, since A'B' is perpendicular to CLc, ∠BCA' = ∠ACB'. The same is true of the pairs of angles at vertices A and B. Let's call this a mirror property. As we know, the orthic triangle of ΔA'B'C' has the mirror property. We'll make use of this observation shortly.
