+0

# Help asap!

+1
56
4

1.In triangle $PQR,$ let $X$ be the intersection of the angle bisector of $\angle P$ with side $\overline{QR}$, and let $Y$ be the foot of the perpendicular from $X$ to side $\overline{PR}$. If $PQ = 9,$ $QR = 10,$ and $PR = 17,$ then compute the length of $\overline{XY}$.

2.The vertices of triangle $ABC$ lie on the sides of equilateral triangle $DEF$, as shown. If $CD = 5$, $CE = BD = 2$, and $\angle ACB = 90^\circ$, then find $AE$. [asy] unitsize(0.6 cm); pair A, B, C, P, Q, R; P = (0,0); Q = (5,0); R = 5*dir(60); B = 2*dir(60); C = (3,0); A = intersectionpoint( C--rotate(-90,C)*B , R--Q ) ; draw(A--B--C--cycle); draw(P--Q--R--cycle); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, S); label("$D$", P, SW); label("$E$", Q, SE); label("$F$", R, N); [/asy]

3.In triangle $ABC$, point $X$ is on side $\overline{BC}$ 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 $\mathcal{R}$ be the circle centered at $(0,0)$ with radius $10.$ The lines $x = 6$ and $y = 4$ divide $\mathcal{R}$ into four regions $\mathcal{R}_1$, $\mathcal{R}_2$, $\mathcal{R}_3$, and $\mathcal{R}_4$. Let $[\mathcal{R}_i]$ denote the area of region $\mathcal{R}_i$. If $[\mathcal{R}_1] > [\mathcal{R}_2] > [\mathcal{R}_3] > [\mathcal{R}_4],$then find $[\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4]$.

Jun 12, 2023

#1
0

1.  Let PX=x and QR=17−x.

[asy] import olympiad; unitsize(1 cm);

pair P, Q, R, X, Y;

P = (0,0); Q = (10,0); R = (17,3); X = extension(P,Q,R,(17,0)); Y = foot(X,P,R);

draw(P--Q--R--cycle); draw(P--X,dashed); draw(X--Y);

label("P", P, SW); label("Q", Q, SE); label("R", R, NW); label("X", X, NE); label("Y", Y, S); label("9", (P + Q)/2, S); label("17−x", (Q + R)/2, S); label("x", (P + X)/2, E); [/asy]

By the Angle Bisector Theorem, [\frac{PX}{QR} = \frac{QX}{PR}.]Substituting in known values, we get [\frac{x}{17 - x} = \frac{9}{17}.]Solving for x, we find x=19153​.

Since XY is the altitude from X to PR, we have [XY^2 = PR \cdot PX - PX^2 = 17 \cdot \frac{153}{19} - \left( \frac{153}{19} \right)^2 = \frac{170}{19}.]Therefore, XY = sqrt(170)/19.

Jun 12, 2023
#2
0

Sorry, but that's wrong :(

Jun 12, 2023
#3
0

2. By the cosine law, AE = 12/5.

Jun 13, 2023
#4
0

Nope :(

Jun 13, 2023